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1
BATRA COACHING CENTR
CENTRE
1326/14, Arjun Nagar, Rohtak, : 9416124333
PAIRS OF LINEAR EQUATIONS IN TWO VARIABLES
Linear Equation in Two Variables
An equation of the type ax + by = c, where a, b, c are real numbers (a ≠ 0, b ≠ 0), is called
a linear equation in two variables.
x y
e.g.
(i)
2x + y = 9
(ii)
x – 3y = 8
(iii)
− =5
(iv)
3m + 2n = 2
4 3
Examples.
x = 4 and y = 1 is a solution of the equation 2x – y = 7, because on substituting
x = 4 and y = 1 in the equation, we have
2×4–1=7
8 – 1 = 7 ⇒ 7 = 7, which is true.
Similarly, x = 5 and y = -11 is a solution of the equation x – 3y = 8.
Geometrically, it means that the point (4, 1) lies on the line represented by the equation
2x – y = 7; and the point (5, -1)
1) lies on the line represented by the equation x – 3y = 8.
Thus, every solution of an equation is a point on the line representing the equation. The
converse is also true.
solutions.
NOTE:: A linear equation in two variables has infinite number of solution
In other words, a linear equation in two variables does not have a unique solution.
We can display many solutions of a given equation in the form of a table as shown below:
For equation 2x – y = 8:
Table of Solutions
x 4 3 0 5 6
y 0 -2 -8 2 4
SIMULTANEOUS LINEAR EQUATION IN TWO VARIABLES
A pair of linear equations in two variables is said to form a system of simultaneous linear
equations. e.g.
3 2
2u + 5v + 1 = 0
(iii)
(iv)
2a + b – 1 = 0
(i)
x + 2y = 3
(ii)
+ =9
x y
2x – y = 5
u – 2v + 9 = 0
1 1
− =5
x y
a + b + 5 = 0.
Solution
A pair of values of the variables x and y satisfying each one of the equations in a
given system of two simultaneous linear equations in x and y is called a solution of the system.
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Example.1
equations.
Show that x = 2, y = 1 is a solution of the system of simultaneous linear
3x – 2y = 4
2x + y = 5.
The given system of equations is
Solution
3x – 2y = 4
2x + y = 5
Putting x = 2 and y = 1 in equation (i), we have
LHS = 3 × 2 – 2 × 1 = 6-22 = 4 = RHS
Putting x = 2 and y = 1 in equation (ii), we have
LHS = 2 × 2 + 1 × 1 =4+1=
4+1= 5 = RHS
Thus, x = 2 and y = 1 satisfy both the equations of the given system.
Hence, x =2, y = 1 is a solution of the given system.
Example 2
Show that x = 2, y = 1 are solutions of the system of equations
3x – 2y = 4
6x – 4y = 8.
The given system of equation is
Solution
3x – 2y = 4
……….. (i)
6x – 4y = 8
………… (ii)
Putting x = 2 and y = 1 in equation (i) and (ii) respectively, we get
LHS = 3 × 2 – 2 × 1 = 4 = RHS
LHS = 6 × 2 – 4 × 1 = 8 = RHS
So, x = 2, y = 1 is a solution of the given of equations.
Consistent System A system of simultaneous linear equations is said to be consistent, if it
has at least one solution.
In-Consistent System A system of simultaneous linear equation is said to be in-consistent,
in
if it
has no solution.
Pairs of Linear Equations
Two linear equations in the same two variables are said to form a pair of linear equations if
both the equations are considered at the same time, i.e., the two equations are either joined by
the conjunction ‘and’ or clubbed together.
Solving a Pair of Linear Equations Graphic
Graphically
Suppose, we have a pair of linear equations, say
ax + by = c, (a ≠ 0, b ≠ 0 )

ex + fy = g , (e ≠ 0, f ≠ 0 ).
a1 x + b1 y = c1
a2 x + b2 y = c2
Then, we can draw the graph of each one of the given equations. Let the graph lines be l and m.
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Subsequently, the following situations shown in the figures that follow may airse.
In case (i), the two graph lines l and m intersect at a point, say P. Let its coordinates be (h, k). So
we say that the pair has a unique solution, namely x = h and y = k. In other words, the system is
consistent.
In case (ii), the two graph lines l and m are parallel, i.e., they do not meet at all. So, we say that
the system has no solution, i.e., there is no common solution of the two given equations. In
other words, the system is inconsistent.
In case (iii), the two graph lines l and m are coinc
coincident,
ident, i.e., every point of l is also a point of m,
or every solution of the first equation is also a solution of the second equation. So, we say that
the system has infinite number of solutions. Hence, the system is consistent.
Examples
owing pair of linear equations graphically:
Q.1
Solve the following
4 x − y = 5

 x− y =5
Sol.
4x – y = 5
⇒
y = 4x – 5.
When x = 0, y = 4 × 0 – 5 = 0 – 5 = -5
When x = 1, y = 4 × 1 – 5 = 4 – 5 = -1.
These values can be displayed in the form of a table as follows:
x 0 1
y -5 -1
5) and B(1, -1)
1) on a graph paper. We join AB and extend it on
Now, we plot the points A(0, -5)
both sides to obtain the line l. The line l is the graph of the equation 4x – y = 5.
Further,
x+y=5⇒ y=5–x
When
x = 0,
y=5–0=5
When
x = 2,
y = 5 – 2 = 3.
These values can be tabulated as follows:
x 0 2
y 5 3
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Now, we plot the point C(0, 5) and D(2, 3) on the same graph paper. On joining CD and
extending it on both sides, we get a line, say m.
On looking at the graph, we find that the two graph lines intersect at a point D(2, 3).
Hence, x = 2 and y = 3 is the required solution. In other words, the given pair of equations is
consistent with a unique solution x = 2, y = 3.
ILLUSTRATIVE EXAMPLES
Ten students of class X to
took
ok part in Mathematics quiz. If the number of girls is 4 more
Q.1
than the number of boys. Represent this situation algebraically and graphically.
Sol.
Let the number of girls be x and the number of boys be y.
It is given that total ten students took part in tthe quiz.
∴
Number of girls + Number of boys = 10
i.e.
x + y = 10
……(i)
It is also given that the number of girls is 4 more than the number of boys.
∴
Number of girls = Number of boys + 4
i.e.
x=y+4
or,
x–y=4
……(ii)
5
6
Thus, two solutions of equation (i) are
x
10
0
y
0
10
Similarly, two solutions of equation (ii) are:
x
4
0
y
0
-4
Now, we plot the points A(10, 0), B (0, 10), P (4, 0) and Q (0, -4)
4) corresponding to these
solutions on the graph paper and draw the lines AB and PQ representing the equations
x + y = 10 and x – y = 4 as shown in Fig. 3.4
We observe that the two lines rep
representing
resenting the two equations are intersecting at the point
(7,3).
Q.2
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, he buys another
bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically
and geometrically.
Sol.
Let the price of a bat be Rs x and that of a ball be Rs. Y.
It is given that 3 bats and 6 balls are bought for Rs. 3900.
∴
3x + 6y = 3900
…….(i)
It is also given that one bat and 3 balls of the same kind cost Rs 1300.
∴
x + 3y = 1300
…….(ii)
3x + 6y = 3900
When y = 0, we have
3900
= 1300
3x + 0 = 3900 ⇒ x =
3
When x = 0, we have
3900
0 + 6y = 3900 ⇒ y =
= 650
6
Thus, two solutions of equation (i) are:
x
1300
0
y
0
650
We have,
x + 3y = 1300
When y = 100, we have
x + 300 = 1300 ⇒ x = 1000
When x = 100, we have
100 + 3y = 1300 ⇒ 3y = 1200 ⇒ y = 400
Thus, two solutions of equation (ii) are:
x
1000
100
7
y
100
400
We observe that the two lines representing the two equations are intersecting at the point A
(1300, 0).
Q.3
Romila went to a stationary stall and purchased 2 pencils and 3 erasers for Rs. 9. Her
friend Sonali saw the new variety of pencils and erasers with Romila, and she also
bought 4 pencils and 6 erasers of the same kind for Rs 18. Represent th
this situation
algebraically and graphically.
Sol.
Let the cost of 1 pencil be Rs. X and that of one eraser be Rs. Y.
It is given that Romila purchased 2 pencils and 3 erasers for Rs. 9.
∴
2x + 3y = 9
It is also given that Sonali purchased 4 pencils and 6 er
erasers for Rs. 18.
∴
4x + 6y = 18
Algebraic Representation: The algebraic representation of the given situation is
2x + 3y = 9
….…(i)
4x + 6y = 18
…….(ii)
We have,
2x + 3y = 9
Putting x = -3, we get
-6 + 3y = 9 ⇒ 3y = 15 ⇒ y = 5
Putting x = 0, we get
0 + 3y = 9 ⇒ y = 3
Thus, two solutions of 2x + 3y = 9 are:
x
-3
0
8
y
5
3
We have,
4x + 6y = 18
Putting x = 3, we get
12 + 6y = 18 ⇒ 6y = 6 ⇒ y = 1
Putting x = -6, we get
-24 + 6y = 18 ⇒ 6y = 42 ⇒ y = 7
Thus, two solutions of 4x + 6y = 18 are
x
3
-6
y
1
7
We find that both the lines AB and PQ coincide.
∴ System has infinite solutions.
Q.4
The path of a train A is given by the equation x + 2y – 4 = 0 and the path of another train
B is given by the equation 2x + 4y – 12 = 0. Represent this situation graphically.
Sol.
The paths of two trains are given by the following pair of linear equations.
x + 2y – 4 = 0
……(i)
2x + 4y – 12 = 0
……(ii)
x + 2y – 4 = 0
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Putting y = 0, we get
x+0–4=0⇒x=4
Putting x = 0, we get
0 + 2y – 4 = 0 ⇒ y = 2
Thus, two solutions of equation x + 2y – 4 = 0 are:
x
y
4
0
0
2
We have,
2x + 4y – 12 = 0
Putting x = 0, we get
0 + 4y – 12 = 0 ⇒ y = 3
Putting y = 0, we get
2x + 0.12 = 0 ⇒ x = 6
Thus, two solutions of equation 2x + 4y – 12 = 0 are:
x
0
6
y
3
0
We observe that the lines are parallel and they do not intersect any where.
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Q.5
Solve the following pair of equations graphically:
 2 x + 7 y = 11


35
5 x + 2 y = 25
Sol.
11 − 2 x
7
11 − 2 × 2 11 − 4
=
=1
When x = 2
y=
7
7
11 − 2 × (− 5) 11 + 10 21
=
=
= 3.
When x = -5,
y=
7
7
7
(Choose only such values of x which would give the values of y as integers and not
rational numbers.)
The above solutions can be displayed as follows:
x
2
-5
y
1
3
2x + 7y = 11
⇒
y=
2
(25 − 5 x ) ⇒ y = 2 (5 − x ).
35
7
2
2
y = (5 + 2 ) = × 7 = 2
When x = -2,
7
7
2
2
y = (5 − 5) = × 0 = 0.
When x = 5,
7
7
These values can be displayed as follows:
x
-2
5
y
2
0
Further, 5x +
35
y = 25
2
⇒
y=
Clearly, the graph lines are parallel lines. As they do not have any common point, the
two equations have no common solution. Hence, the given pair of equations is inconsistent.
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ASSIGNMENT – I
Solve the following pair of equation graphically:
Q.1
 2x − y = 7

6 y − 3x = 21
Q.2
The sides of a triangle are given by the equations y – 2 = 0, y = 3x – 7 and 2y + x = 0. Find
graphically the coordinates of the vertices of the triangle.
Q.3
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and
play Hoopla (a game in which you throw a ri
ring on the
he items kept in the stall, and if the
ring covers any object completely you get it). The number of times she played Hoopla is
half the number of rides she had on the Giant Wheel. Each ride costs Rs 3, and a game
of Hoopla costs Rs 4. If she spent Rs 20 in the fair, represent this situation algebraically
and graphically.
The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another
Q.4
train B is given by the equation 6x + 8 y – 48 = 0. Represent this situation graphically.
c
a b
Q.5
On comparing the ratios 1 , 1 and 1 , and without drawing them, find out whether the
a2 b2
c2
lines representing the following pairs of linear equations intersect at a point, are parallel
or coincide:
(i)
5x – 4y + 8 = 0
(ii)
9x + 3y + 12 = 0
(iii)
6x – 3y + 10 = 0
7x + 6y – 9 = 0
18x + 6y + 24 = 0
2x – y + 9 = 0
(iv)
x – 2y = 2
(v)
3x + y = 2
(vi)
3x – 4y = 1
3x + 5y = 17
6x + 2y = 5
8y – 6y = 4
(vii) y = 2x – 3
(viii) 5x + 2y = 16
2x = y + 3
7.5x + 3y = 24
Q.6
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables
such that the geometrical representation of the pair so formed is:
(i)
intersecting lines
(ii)
parallel lines
(iii)
coincident lines.
Q.7
Represent the following system of linear equations ggraphically.
raphically. From the graph, find the
points where the lines intersect yy-axis:
3x + y – 5 = 0; 2x – y – 5 = 0.
Q.8
Solve the following pair of linear equations graphically:
4x – 5y – 20 = 0
3x + 5y – 15 = 0
Determine the vertices of the triangle forme
formed
d by the lines, representing the above
equations, and the y-axis.
axis.
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Q.9
(i)
(ii)
Q.10
Q.11
(i)
(iv)
Q.12
(i)
(ii)
(iii)
Q.13
(i)
(ii)
(iii)
Q.14
(i)
Q.15
Q.16
Q.17
Q.18
Form the pair of linear equations in the following problems, and find their solutions
graphically:
10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more
than the number of boys, find the number of boys and girls who took part in the quiz.
5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs.
46. Find the cost of one pencil and a pen.
Find the dimensions of a garden whose length is 4m more than its width and half the
perimeter is 36 m by first writing them algebraically and solving the pair of linear
equations graphically.
Solve each of the following pair of linear equations graphically:
2x + 3y = 8
(ii)
x – 2y = 4
(iii)
2x – 3y = 1
x + 4y = 9
x–y=3
4x – 3y + 1 = 0
x+y=3
(v)
3x – y = -4
2x + 5y = 12
5x + y = -4
Solve the following pairs of linear equations graphically:
2x – 3y = -4; x + 2y = 5
Also, find the points where these lines intersect the xx-axis.
2x – y = 4; 3y – x = 3
Also, find the points where these lines intersect the yy-axis.
2x + 3y = 12; x – y = 1
Shade the region (area) between the two lines and the xx-axis.
Find graphically the coordinates of the vertices of a triangle whose sides have the
equations:
y = x, y = 0 and 2x + 3y = 30
2y – x = 8, 5y – x = 14 and y – 2x = 1
y = x, 3y = x and x + y = 8
For each of the following pairs of equations, draw the graph lin
lines
es and calculate the area
bounded by the graph lines and the xx-axis:
4x – 3y + 4 = 0
(ii)
2x + y = 6
(iii)
x+y=5
4x + 3y – 20 = 0
2x – y + 2 = 0
2x – y + 2 = 0
Draw the graphs of the following equations: 2x – 3y + 6 = 0, y = 2 and 2x + 3y= 18.
Find the vertices of the triangle so obtained. Also, find the area of the triangle.
The sum of two numbers is 40. If the smaller number is doubled, it becomes 14 more
than the greater number. Find the numbers.
The combined railway fare for a journey undertaken by a family of 4 members travelling
in 3-tier
tier coach and a family of 3 members travelling in 22-tier
tier coach is Rs. 5100. The total
fare would have been Rs. 300 more if the first family had 1 member less while the
second had 1 member m
more.
ore. What was the fare for a couple for the same journey in 22
tier?
Two persons A and B together can do a work in 10 days. The same work can be finished
in 18 days if A works alone for 12 days and B works alone for 6 days. Find the time
required for each
ach one of them to finish the work if they work together.
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14
ANSWERS ASSIGNMENT – I
Infinite number of solutions. 2.
1.
(2, -1), (3, 2) and (-4, 2).
3.
x – 2y = 0
3x + 4y = 20
Parallel lines
Inconsistent
5.(i) Intersecting lines
(ii)
Coincident lines
(iii)
(iv)
Unique solution
(v)
No unique solution
(vi)
(vii) Consistent
(vi)
Consistent
6.(i) x + 2y – 4 = 0
(ii)
4x + 6y – 12 = 0
(iii)
4x + 6y – 16 = 0
7.
A(0,5) P(0, -5)
8.
A(0, 3), B(0, -4) and C(5, 0).
9(i)
(ii)
The cost of one pencil is Rs. 3 and the cost of one pen is Rs. 5.
10.
20m × 16m.
11.(i) x = 1, y = 2
(ii) x = 2, y = -1
(iii) x = -1, y = 1
(iv) x = 1, y = 2 (v) x = -1, y = 1
12.(i) x = 1, y = 2; the lines intersect the xx-axis at (-2, 0) and (5, 0) respectively.
(ii)
x = 3, y = 2; the lines intersect the yy-axis at (0, -4)
4) and (0, 1) respectively.
(iii)
x = 3, y = 2; the lines intersect the xx-axis at (6, 0) and (1, 0) respectively.
13.(i) (15, 0), (0, 0), (6, 6) (ii)
(-4, 2), (1, 3), (2, 5)
(iii)
(0, 0), (4, 4), (6, 2).
14.(i) 12 sq units
(ii)
12 sq units
(iii)
8 sq units
(iv)
12 sq units
15.
(3, 4), (0, 2), (6, 2); Area = 6 sq units
16.
22, 18
17.
Rs. 1800
18.
15 days, 30 days.
ALGEBRAIC METHODS OF SOLVING SIMULTANEOUS LINEAR EQUATIONS IN TWO VARIABLES
The most commonly used algebraic methods of solving simultaneous linear equations in
two variables are:
Method of elimination by substitution
(i)
(ii)
Method of elimination by equating the coefficients.
(iii)
Method of cross-multiplication.
multiplication.
METHOD OF ELIMINATION BY SUBSTITUTION
ALGORITHM
STEP I Obtain the two equations. Let the equations be
a1x + b1y + c1 = 0
……(i)
and, a2x + b2y + c2 = 0
…….(ii)
STEP II Choose either of the two equations, say (i), and find the value of one variable, say y, in
terms of the other, i.e. x.
STEP III Substitute the value of y, obtained in step II, in the other equation i.e. (ii) to get an
equation in x.
STEP IVSolve
Solve the equation obtained in step III to get the value of x.
STEP V Substitute the value of x obtained in step IV in the ex
expression
pression for y in terms of x
obtained in step II to get the value of y.
STEP VI
The values of x and y obtained in steps IV and V respectively constitute the
solutions of the given system of two linear equations.
Following solved examples will illustrate the above algorithm
ILLUSTRATIVE EXAMPLES
Solve the following system of equations by using the method of substitution:
Q.1
(i)
3x – 5y = -1
(ii)
x + 2y = -1
15
x–y=-1
2x – 3y = 12
Sol.(i) The given system of equation is
3x – 5y = -1
…..(i)
x – y = -1
…..(ii)
From (ii), we get
y=x+1
Substituting y = x + 1 in (i), we get
3x – 5 (x + 1) = -1
⇒
-2x – 5 = -1
⇒
-2x = 4 ⇒ x = -2
Putting x = -22 in y = x + 1 we get y = -1.
Hence, the solution of given system of equations is x = -2, y = -1.
(ii)
The given system of equations is
x + 2y = -1
……(i)
2x – 3y = 12
……(ii)
From equation (i), we get
x = -1 – 2y
Substituting x = -1 – 2y in equation (ii) we get
⇒
2(-1 – 2y) – 3y = 12
⇒
-2 – 4y – 3y = 12
⇒
-7y = 14
⇒
y = -2
Putting y = -2 in x = -1 -2y,
2y, we get
X = -1 – 2 × (-2) = 3
Hence, the solution of the given system of equations is x = 3, y = -2.
Q.2
Solve for x and y by substitution method:
x + 5y = 34
x – 5y = -6
The given pair of equations are
Sol.
x + 5y = 34
…..(1)
x – 5y = -6
…..(2)
From Eq. (1), x = 34 – 5y.
Substituting 34 – 5y for x in Eq. (2), we get
34 – 5y – 5y = -6
⇒
34 – 10y = -6
⇒
10y = 40
⇒
y = 4.
From Eq. (3), x = 34 – 5 × 4 = 34 – 20 = 14
Thus, x = 14 and y = 4 is the required solution.
Check: Substituting
ing x = 14 and y = 4 in Eq. (1) and Eq. (2), we have
L.H.S. of Eq. (1) = 14 + 5 × 4 = 14 + 20 = 34 = R.H.S. of Eq. (1)
L.H.S. of Eq. (2) = 14 – 5 × 4 = 14 – 20 = -6 = R.H.S. of Eq. (2).
Q.3
Solve for x and y by substitution method:
2x + 3y = 23
16
5x – 20 = 8y
The pair of linear equations on rearrangement becomes
Sol.
2x + 3y = 23
….(1)
5x – 8y = 20
…..(2)
23 − 3 y
Substitution Method: From Eq. (1), x =
.
2
23 − 3 y
 23 − 3 y 
Substituting
for x in Eq. (2), we get 5 × 
 − 8 y = 20
2
 2 
⇒
115 – 15y – 16y = 40
75
⇒
31y = 75 ⇒ y =
31
75
Substituting y =
in Eq. (5), we get
31
1  713 − 225 
1
75 
x =  23 − 3 ×  ⇒ x = 

31 
2
31
2

1  488  244
⇒
x= 
=
2  31  31
244
75
Thus, x =
and y =
is the required solution.
31
31
Q.4
Solve for x and y:
x y
+ =1
6 4
3x x − y 7
−
=
4
2
4
The given pair of linear equations are
Sol.
x y
+ =1
….(1)
6 4
3x x − y 7
−
…..(2)
=
4
2
4
Multiplying Eq. (1) by 12 (The LCM of 6 and 4), we get
2x + 3y = 12
……(3)
Multiplying Eq. (2) by 4 (The LCM of 2 and 4), we get
3x – 2(x – y) = 7
⇒
3x – 2x + 2y = 7
⇒
x + 2y = 7
……(4)
From Eq. (4), x =7-2y
2y for x in Eq. (3
(3), we get
Substituting 7-2y
2(7-2y)+3y=12 ⇒14−4
17
 12 − 3 y 
2
 + 4 y = 14
 2 
⇒
12 – 3y + 4y = 14
⇒
12 + y = 14
⇒
y=2
Substituting y = 2 in Eq. (5), we get 2x + 3 × 2 = 12
⇒
2x + 6 = 12
⇒
x = 6 ÷ 2 = 3.
Thus, x = 3 and y = 2 are the required values.
Check: Substituting x = 3 and y = 2 in Eq. (1) and Eq. (2), we find that
x y 3 2 1 1
L.H.S. of Eq. (1) = + = + = + = 1 = R.H .S . of Eq. (1)
6 4 6 4 2 2
3x x − y 3 × 3 3 − 2 9 1 7
L.H.S. of Eq. (2) =
−
=
−
= − = = R.H .S . of Eq. (2)
4
2
4 2 4
4
2
Q.5
Solve the following systems of equations by using the method of substitution:
x y
2x y
+ = 2 and − = 4
a b
a b
Sol.
The given system of equation is
2x y
+ =2
a b
x y
− =4
a b
From equation (i), we get
y
2x
2x 

=2−
⇒ y = b 2 − 
b
a
a 

2x 

Substituting y = b  2 −  in equation (ii), we get
a 

x b
2x 
− 2 −  = 4
a b
a 
x
2x
⇒
−2+
=4
a
a
3x
⇒
=6
a
⇒
3x = 6a
⇒
x = 2a
Putting x = 2a in equation (i), we get
18
y
y
= 2 ⇒ = −2 ⇒ y = −2b
b
b
Hence, the solution of the given system of equations is x = 2a, y = -2b.
Following algorithm explains the procedure.
Method of elimination
limination by equating the coefficients.
ALGORITHM
STEP I Obtain the two equations.
STEP II Multiply the equations so as to make the coefficients of the variable to be eliminated
equal.
STEP III Add or substract the equation obtained in step II according aass the terms having the
same coefficients are opposite or of the same sign.
STEP IVSolve
Solve equation in one variable obtained in step III.
STEP V Substitute the value found in step IV in any one of the given equations and find the
value of the other variable.
The values of the variables in steps IV and V constitute the solution of the given system
of equations.
Following examples will illustrate the above algorithm.
ILLUSTRATIVE EXAMPLES
TYPE I SOLVING SIMULTANEOUS LINEAR EQUATION IN TWO VARIABLES
Q.1
Solve the following system of linear equation by using the method of elimination by
equation the coefficients:
(i)
3x + 2y = 11
(ii)
8x + 5y = 9
2x + 3y = 4
3x + 2y = 4
Sol.(i) The given system of equations is
3x + 2y = 11
…..(i)
2x + 3y = 4
…..(ii)
Let us eliminate y from the given equations. The coefficients of y in the given equations are 2
and 3 respectively. The l.c.m. of 2 and 3 is 6. So, we make the coefficients of y equal to 6 in the
two equations.
Multiplying (i) by 3 and (ii) by 2, we ge
get
9x + 6y = 33
……(iii)
4x + 6y = 8
……(iv)
Subtracting (iv) from (iii), we get
5x = 25 ⇒ x = 5
ting x = 5 equation in (i), we get
Substituting
15 + 2y = 11 ⇒ 2y = -4 ⇒ y = -2
Hence, the solve of the given system of equation is x = 5, y = -2
(ii)
The given
ven system of equation is
8x + 5y = 9
……(i)
3x + 2y = 4
……(ii)
4+
19
Let us eliminate x from the given equations. The coefficients of x in the given equations are 8
and 3 respectively. The l.c.m. of 8 and 3 is 24. So, we make both the coefficients equal to 24.
Multiplying both sides of equation (i) by 3 and equation (ii) by 8, we get
24x + 15y = 27
24 + 16y = 32
Subtracting (iv) from (iii), we get
-y = -5 ⇒ y = 5
Putting y = 5 in (i), we get
8x + 25 = 9 ⇒ 8x = -16 ⇒ x = -2
Hence, the solution of the given
iven system of equations is x = -2, y = 5.
20
Q.2
Solve for x and y by Elimination Method.
ax + by = 3ab
a2x + b2y = a + b
Sol.
Given equation are
…..(1)
ax + by = 3ab
2
2
a x+b y=a+b
……(2)
Multiplying Eq. (1) by a, we get
a2x + aby = 3a2b
……(3)
Subtracting Eq. (2) from Eq. (3), we get
(ab – b2) y = 3a2b – a – b
3a 2b − a − b
⇒
y=
ab − b2
Multiplying Eq. (1) by b, we get
abx + b2y = 3ab2
Subtracting Eq. (2) from Eq. (4), we get
(ab – a2) x = 3ab2 – a – b
3ab2 − a − b
⇒
x=
ab − a 2
3ab2 − a − b
3a 2b − a − b
Thus, x =
and y =
.
ab − a 2
ab − b 2
Q.3
Solve for x and y:
b
a
x + y = a2 + b2
a
b
x + y = 2ab
Sol.
Given equation are
b
a
…..(1)
x + y = a2 + b2
a
b
x + y = 2ab
…..(2)
b
Multiplying Eq. (2) by , we get
a
b
a
x + y = 2b 2
a
b
Subtracting Eq. (3) from Eq. (1), we get
 a 2 − b2 
a b
2
2
 y = a 2 − b2

=
−
y
a
b
−
⇒


b a
 ab 
Q.4
⇒
y = ab
From Eq. (2), x = 2ab – y = 2ab – ab = ab
Hence, x = ab and y = ab is the required solution.
Solve for x and y by Substitution Method:
px + qy = 0
21
Sol.
lx + my = n
Given equations are
px + qy = 0
lx + my = n
From Eq. (1), we have
p
y=− x
....(3)
q
p
Substituting y = − x in Eq. (2), we get
q
⇒
nq
lq − pm
nq
Substituting
in Eq. (3), we get
x=
lq − pm
np
y=−
lq − pm
nq
np
Thus, x =
,y=−
lq − pm
lq − pm
Solve for x and y:
148x + 231 y = 527
231x + 148y = 610
Given equation are
148x + 231 y = 527
……(1)
231x + 148y = 610
……(2)
Adding Eq. (1) and Eq. (2), we get
379x + 379 y = 1137
⇒
x+y=3
…….(3)
Subtracting Eq. (1) from Eq. (2), we get
83x – 83y = 83
⇒
x–y=1
…….(4)
Adding Eq. (3) and Eq.. (4), we get
2x = 4
⇒
x=2
Substituting x = 2 in Eq. (3), we get
Y=1
Hence, the required solution is x = 2 and y = 1.
⇒
Q.5
Sol.
 p 
lx + m − x  = n
 q 
 pm 
x = n
 l −
q 

x=
22
TYPE II SOLVING A SYSTEM OF EQUATIONS WHICH IS REDUCIBLE TO A SYSTEM OF
SIMULTANEOUS LINEAR EQUATIONS
Q.6
Solve the followingg system of equations:
1 1
− = −1
2x y
1 1
+
= 8, where x ≠ 0, y ≠ 0.
x 2y
1
1
Sol.
Taking = u and = v, the given equations become
x
y
u
− v = −1⇒ u − 2v = −2
……(i)
2
v
and
u + = 8 ⇒ 2u + v + 16
…….(ii)
2
Let us eliminate u from equation (i) and (ii). Multiplying equation (i) by 2, we get
2u – 4v = -4
…..(iii)
2u + v = 16
……(iv)
Subtracting (iv) from (iii), we get
-5v = -20 ⇒ v = 4
Putting v = 4 in equation (i), we get
u – 8 = -2 ⇒ u = 6
1 1
1 1
Hence, x = = and y = =
u 6
v 4
1
1
So, the solution of the given system of equation is x = , y = .
6
4
2 2 1
Q.7
Solve: +
=
x 3y 6
3 2
+ =0
x y
and hence find ‘a’ for which y = ax – 4.
1
1
Sol.
Taking = u and = v. The given system of equation
equations become
x
y
2
1
2u + v =
3
6
⇒
12u + 4v = 1……..(i)
and, 3u + 2v = 0 ……..(ii)
Multiplying equation (ii) by 2 and subtracting from equations (i), we get
1
6u = 1 ⇒ u =
6
1
Putting u = in (i), we get
6
23
1
4
1
1
Hence, x = = 6 and y = = −4
u
v
So, the solution of the given system of equations is x = 6, y = -4
Putting x = 6, y = -44 in y = ax – 4, we get
-4 = 6a – 4 ⇒ a = 0
Solve for x and y:
30
44
+
= 10
x− y x+ y
40
55
(x − y ≠ 0, x + y ≠ 0).
+
= 13,
x− y x+ y
1
1
Let
= p and
= q. Then the given equations become
x− y
x+ y
30p + 44q = 10
…..(1)
40p + 55q = 13
……(2)
Multiplying Eq. (1) by 4 and Eq. (2) by 3, we get
120p + 176q = 40
……(3)
120p + 165q = 39
……(4)
Subtracting Eq. (4) from Eq. (3), we get
11q = 1
1
⇒
q=
11
1
1
or
=
x + y 11
1
Substituting q =
in Eq. (3), we get
11
1
120 p + 176  = 40
 11 
⇒
120p + 16 = 40
⇒
120p
p = 24
1
⇒
p=
5
1
Or
…..(6)
=5
x− y
Eq. (5) and Eq. (6) further give us
x + y = 11
and
x–y=5
Solving these two for x and y, we get
x = 8 and y = 3.
2 + 4v = 1 ⇒ v = −
Q.8
Sol.
24
Q.9
Sol.
Q.10
Sol.
Solve for x and y:
1
5
3
+
=−
2(x + 2 y ) 3(3x − 2 y )
2
61
5
3
−
= , where x + 2y ≠ 0 and 3x – 2y ≠0.
4(x + 2 y ) 5(3x − 2 y ) 60
1
1
Let
= u,
= v. Then the given equations become
x + 2y
3x − 2 y
3
u 5v
+ =−
2 3
2
5u 3v 61
− =
4 5 60
⇒
3u + 10v = -9
75u – 36v = 61
Now, Eq. (2) – 25 × Eq. (1)
⇒
(75u – 36v) – 25 ( 3u + 10v) = 61 – 25 × (-9)
⇒
-286v
286v = 286 ⇒ v = -1
From Eq. (1),
3u = -9 – 10v = -9 + 10 = 1
1
⇒
u=
3
Now, these imply
x + 2y = 3
3x – 2y = -1
Adding these equations, we get
1
4x = 2 ⇒ x =
2
1
5
Also, x + 2y = 3 ⇒ 2y = 3 - ⇒ y = .
2
4
1
5
Hence x = and y = is the required solution of the given equations.
2
4
Solve:
x + y = 5xy
3x + 2y = 13xy
(x ≠ 0, y ≠0).
Dividing both the given equations by xy
(Note: x ≠ 0, y ≠0 ⇒ xy ≠ 0), we have
1 1
……(1)
+ =5
y x
3 2
……(2)
+ = 13
y x
1
1
Let = u,
= v, Eqs. (1) and (2) become
x
y
25
Q.11
Sol.
u+v=5
…..(3)
2u + 3v = 13
…..(4)
Now, Eq. (4) – 2 × Eq. (3)
⇒ 2u + 3v – 2(u + v) = 13 – 2 × 5
1
⇒ v=3⇒x= .
3
From Eq. (3), u = 5 – v = 5 – 3 = 2
1
⇒u=2⇒x= .
2
1
1
Thus, x = , y = is the solution of given equations.
2
3
Solve:
xy
6
=
x+ y 5
xy
= 6, where x + y ≠ 0 and y – x ≠0.
y−x
x ≠0 y ≠0 and x ≠y.
∴ The given equations imply
5xy = 6(x + y)
Xy = 6 (y – x)
Dividing by xy (Here, xy ≠ 0), we have
1 1
6 +  = 5
 y x
1 1
6 −  = 1
x y
1
1
Let = u , = v, these equations become
x
y
6u + 6v = 5
6u – 6v = 1
Adding these equations, we get
1
12 u = 6 ⇒ u = ⇒ x = 2
2
Also, 6v = 5 – 6u = 5 – 3 = 2
1
⇒
v = ⇒ y = 3.
3
Hence, x = 2, y = 3 is the required solution.
TYPE IV
EQUATION OF THE FORM
a1x + b1y + c1 z= d1
a2x + b2y + c2 z= d2
26
a3x + b3y + c3 z= d3
To solve the above type of equation, following algorithm may be used.
ALGORITHM
STEP I Take any one of the three equations.
STEP II Obtain the value of one of the variable, say z from it.
STEP III Substitute the value of z obtained in step II in the remaining two equations to obtain
two linear equations in x, y.
STEP IVSolve
Solve the equations in x, y obtained in step III by elimination method.
STEP V Substitute the values of x, y obtained in Ste
Step IV and step II to get the value of z.
Following examples illustrate the above procedure.
Q.12 Solve: 2x – y = 4
y-z=6
x – z = 10
Sol.
We have,
2x – y = 4
y–z=6
x – z = 10
From equation (iii), we get z = x = 10
Substituting the value of z in equation (ii), we get
Y – (x – 10) = 6
⇒
-x + y = -4
Adding equations (i) and (iv), we get
x= 0
Putting x= 0 equation
on in (i) and (iii) we get
Y = -4 and z = -10
Hence, x = 0, y = -4, z = -10
10 is the solution of the given system of equations.
ASSIGNMENT
Solve the following pairs of equations
Q.1
(i)
x+y=6
(ii)
5x + 4y = 4
(iii)
11x + 15y + 23 = 0
x–y=2
x – 12 y = 20
7x – 2y – 20 = 0
x y
4
=
(iv)
3x – 7y + 10 = 0
(v)
(vi)
+ 5y = 7
3 6
x
4x
3
+ 2y = 4
Y – 2x – 3 = 0
+ 4y = 5
x
3
Q.2
Solve the following pairs of linear equations:
x y
(i)
7(y + 3) – 2 (x + 2) = 14
(ii)
+ =5
(iii)
x – y = 0.9
7 3
x y
11
4 (y – 2) + 3 (x – 3) = 2
− =6
=2
x+ y
2 9
3
3
15 2
(iv)
(v)
(vi)
+ 3y = 8
2x − = 9
+ = 17
x
y
x y
27
7
=2
y
y
x+ =4
2
x
+ 2y = 5
3
3x +
(vii)
1 1 36
+ =
x y 5
6
− 4 y = −5
x
28
Q.3
(i)
(iv)
(vii)
Q.4
(i)
(iv)
Q.5
(i)
(iv)
Solve the following pairs of linear equations:
y+7
3x −
+ 2 = 10
(ii)
2x − 3 y = 0
11
x + 11
2y +
= 10
3x − 8 y = 0
7
1
1
x+ y
(v)
+
=2
=2
2x 3y
xy
1
1 13
x− y
+
=
=6
3x 2 y 16
xy
2 5
+ =1
x y
60 40
+
= 19, x ≠ 0, y ≠ 0
x
y
Solve the following pairs of linear equations:
1
7 8
2x − y =
(ii)
+ =2
2
x y
5
2 12
=1
+ = 20
x y
2(x + y )
5
1
6
(v)
+
=2
x+ =6
x −1 y − 2
y
6
3
8
−
=1
3x − = 5
x −1 y − 2
y
Solve for x and y:
39x + 41y = 76
99x + 101 y = 499
(ii)
101x + 99y = 501
41x + 39y = 84
152x -378y = -74
-378x + 152y = -604
(iii)
(vi)
(iii)
1
1
+
=3
7x 6 y
1
1
−
=5
2x 3 y
1
1
+
= 12
5x 6 y
1
3
−
= 8, x ≠ 0, y ≠ 0
3x 7 y
x+ y = 7xy
2 3
+ = 17
x y
(vi)
8x – 9y = 6xy
10x + 6y = 19 xy
(iii)
3(2x + y) = 7xy
3(x + 3y) = 11xy
x
.
y
Q.7(i) Solve: 2x + 3y = 11 and x – 2y = -12. Hence, find the value
alue of m for which y = mx + 3.
6
8
(ii)
Solve: 4 x + = 15 and 6 x − = 14. Hence, find ‘λ’ if y = λx – 2.
y
y
2
3
5
4
+
= 13 and
−
= −2.
Q.8
Solve:
x
y
x
y
2
3
17
Q.9
Solve:
+
=
3x + 2 y 3x − 2 y 5
5
1
+
=2
3x + 2 y 3x − 2 y
Q.6
If 2x + y = 35 and 3x + 4y = 65, find the value of
29
Hence, find x + 4y and 5y – 3x.
Q.10 Solve the following pairs of linear equations:
44
30
+
= 10
x+ y x− y
55
40
+
= 13
x+ y x− y
Q.11 Solve the following pairs of linear equations:
x–y+z=4
x – 2y – 2z = 9
2x + y + 3z = 1
Q.12 Solve the following pairs of linear equations:
x–y+z=4
x+y+z=2
2x + y – 3z = 0
Answers
3
1.(i) x = 4, y = 2
(ii)
x = 22, y = −
(iii)
x = 2, y = -3 (iv)
2
3
3
1
(v)
x= ,y=
(vi)
x = , y = −1
4
2
3
2.(i) x = 5, y = 1
(ii)
x = 14, y = 9 (iii)
x = 3.2, y = 2.3 (iv)
1
(v)
x = 5, y =
(vi)
x = 2, y = 2
(vii) x = 3, y = 2
7
1
1
3.(i) x = 0, y = 0
(ii)
x = 3, y = 4
(iii)
x=
(iv)
,y=
14
6
1
89
1
89
x = − , y = (vi)
,y=
(v)
x=
s
(vii)
2
4080
4
1512
3
1
1
4.(i) x= -1, y = 2
(ii)
x = 1, y =
(iii)
(iv)
x= ,y=
2
4
3
5.(i)
x = 3, y = 2
(ii)
x = 3, y = -1
(iii)
6.
3.
7.(i)
x = -2, y = 5 and m = -1
x=
x = 2, y = 2
1
1
x= ,y=
2
3
x = 4, y = 10
x = 4, y = 5
x = 0, y = 0 and x = 1, y =
(ii)
1
1
,y=
9.
x = 1, y = 1, x + 4y = 5 and 5y – 3x = 2
4
9
11.
x = 3, y = -2, z = -1
12.
x = 2, y = -1, z = 1.
METHOD OF CROSS-MULTIPLICATION
MULTIPLICATION
Q.1
Solve the each of the following systems of equations by using
multiplication:
(i)
x+y=7
(ii)
2x + 3y = 17
(iii)
5x + 12y = 7
3x – 2y = 6
Sol.(i) The given system of equations is
8.
x = -1,
1, y = 1
3
2
x = 3, y = 2 and λ =
10.
4
3
x = 8, y = 3
the method of cross
cross2x + y – 35 = 0
3x + 4y – 65 = 0
30
x+y=7
5x + 12y = 7
By cross-multiplication, we get
x
y
−1
1
7
1
1
12
7
5
12
x
y
−1
=
=
⇒
7 − 84 35 − 7 12 − 5
x
y −1
⇒
=
=
− 77 28 7
x
y −1
−1
⇒
and
=
=
− 77 7
28 7
77
− 28
x=
and
y=
⇒
7
7
x = 11
and
y = −4
⇒
Hence, the solution of the given system of equations is x = 11, y = -4.
(ii)
The given system of equations is
2x + 3y = 17
3x – 2y = 6
By cross-multiplication, we have
x
y
−1
3
17
2
3
−2
6
3
−2
x
y
−1
=
=
⇒
3 × 6 − 17(− 2) 17 × 3 − 6 × 2 2(− 2) − 3 × 3
x
y
−1
⇒
=
=
18 + 34 51 − 12 − 4 − 9
x
y
−1
=
=
⇒
52 39 − 13
1
1
x
y
⇒
=
and
=
52 13
39 13
52
39
x=
and
y=
⇒
13
13
x=4
and y = 3
⇒
Hence, x = 4, y = 3 is the solution of the given system of equations.
(iii)
The given system of equation is
2x + y = 35
3x + 4y = 65
31
x
1
4
y
−1
35
65
2
1
3
4
x
y
−1
⇒
=
=
65 − 140 105 − 130 8 − 3
x
y
−1
⇒
=
=
− 75 − 25 5
−1
x
y
−1
=
⇒
and
=
− 25 5
− 75 5
75
25
⇒
x=
and
y=
5
5
x = 15
and y = 5
⇒
Hence, the solution of the given system of equation is x = 15, y = 5.
Q.2
Solve the following pair of linear equations by cross
cross-multiplication
multiplication method:
bx ay
− +a+b =0
a b
and bx – ay + 2ab = 0
Sol.
The given equations can be written as
b2x – a2y = -ab(a + b)
and bx – ay = -2ab
By cross-multiplication
multiplication method, we have
y
x
−1
647
48 644
47
44
48
4 647
48
2
2
−a
−ab(a + b)
b
−a 2
−a
− 2ab
b
−a
−1
x
y
=
=
2
2
3
2a b − a b(a + b) − ab (a + b) + 2ab − ab2 + a 2b
y
x
−1
=
=
⇒
2
2
a b[2a − a − b] − ab [a + b − 2b] − ab[b − a]
1
x
y
=
⇒
=
2
2
a b(a − b ) − ab (a − b ) ab(b − a )
x
1
y
1
=
=
⇒
and
2
2
a b(a − b) ab(b − a )
− ab (a − b ) ab(b − a )
2
a b(a − b )
− ab2 (a − b )
x=
and y =
⇒
ab(b − a )
− ab(a − b)
⇒
x = -a
and y = b
Thus, x = -aa and y = b is the required solution.
Q.3
Solve the following pairs of linear equations by cross-multiplication
multiplication method:
ax + by = a − b
(i)

bx − ay = a + b
⇒
3
32
2(ax − by ) + a − 4b = 0

2(bx + ay ) + b − 4a = 0
a a
b
b
Sol.(i) Here, 1 = and 1 =
a2 b
b2 − a
a b
∴ 1≠ 1
a2 b2
Therefore, the given pair of equations has a unique solution.
By cross-multiplication method, we have
y
x
−1 48
64
47
44
8 647
4
8 647
b
b
a −b
a
−a
−a
a +b
b
x
y
−1
=
⇒
=
b(a + b ) + a(a − b ) b(a − b) − a(a + b) − a 2 − b2
y
−1
x
⇒
=
=
2
2
2
2
2
ab + b + a − b
ab − b − a − ab − (a 2 + b 2 )
1
x
y
=
= 2
⇒
2
2
2
2
a +b
a + b2
−a −b
1
y
1
x
⇒
= 2
= 2
and
2
2
2
2
2
a +b
a +b
− (a + b ) a + b 2
a 2 + b2
− (a 2 + b 2 )
⇒
and
x= 2
y
=
a + b2
a2 + b2
x=1
and y = -1
⇒
Hence, the unique solution of the given equations is x = 1 and y = -1.
(ii)
2ax + 2by = -a + 4b
2bx + 2ay = 4a – b.
a 2a a
b − 2b − b
Here, 1 =
= and 1 =
=
a2 2b b
b2
a
2a
a1 b1
∴
≠ .
a2 b2
Therefore, the given pair of equations has a unique solution.
By cross-multiplication
multiplication method, we have
x
y
−1
2b
− a + 4b
2a
2b
2a
4a − b
2b
2a
x
y
−1
⇒
= 2
=
2b(4a − b ) − 2a(− a + 4b ) 2b(− a + 4b ) − 2a(4a − b ) 4a − 4b 2
x
y
−1
⇒
=
=
2
2
2
2
2
8ab − 2b + 2a − 8ab − 2ab + 8b − 8a + 2ab 4(a − b 2 )
(ii)
33
x
y
−1
=
=
2
2
2
2
2 a −b
−8 a −b
4 a − b2
−1
−1
x
y
⇒
=
and
=
2
2
2
2
2
2
2
2 a −b
4 a −b
4 a − b2
−8 a −b
− 2 a 2 − b2
8 a 2 − b2
⇒
x=
y
=
and
4 a 2 − b2
4 a 2 − b2
1
x=−
and
⇒
y = 2.
2
1
Hence, x = − and y = 2 is the required solution.
2
Q.4
Solve the following pair of equations by cross
cross-multiplication method:
6(ax + by) = 3a + 2b
6(bx – ay) = 3b – 2a
b
a a
b
Sol.
Here, 1 = and 1 = −
a2 b
b2
a
a1 b1
∴
≠ .
a2 b2
Hence, the given pair of equations has a unique solution.
By cross-multiplication
multiplication method, we have
u
v
1
6447448 647
48 647
7
48
6b
(3a + 2b)
6a
6b
− 6a (3b − 2a)
6b
−6a
x
y
−1
⇒
=
=
6b(3b − 2a ) + 6a(3a + 2b ) 6b(3a + 2b ) − 6a(3b − 2a) − 36a 2 − 36b 2
x
y
−1
⇒
=
=
2
2
2
2
18b − 12ab + 18a + 12ab 18ab + 12b − 18ab + 6a
− 36a 2 − 36b
36 2
− 18a 2 − 18b2 1
− 12a 2 − 12b2 1
x=
and
y
=
=
⇒
=
− 36a 2 − 36b2 2
− 36a 2 − 36b2 3
1
1
Hence, x = and y = is the solution of the given equations.
2
3
Q.5
Solve the following pair of equations by cross
cross-multiplication method:
2
2
2
2
a b
a b ab
− = 0 and
+
= a + b(x ≠ 0, y ≠ 0)
x
y
x
y
1
1
Sol.
Let = u and = v. Then the given equations become
x
y
2
a u – b2u = 0
a2bu + ab2u = (a + b)
a1 a 2 1
b1 − b2
1
= 2 = and = 2 = −
Here,
b2 ab
a
a2 a b b
⇒
(
)
2
(
(
(
) (
(
) (
)
)
)
)
(
(
(
) (
)
)
)
34
a1 b1
≠ .
a2 b2
Thus, the Eqs. (1) and (2) have a unique solution.
By cross-multiplication
multiplication method, we have
u
v
−1
6447448 647
48 647
48
2
2
a
−b 2
−b
0
∴
ab2
( a + b)
a 2b
ab2
u
v
−1
= 3 2
=
⇒
2
2
− b (a + b ) − 0 0 − a (a + b ) a b + a 2b3
u
v
−1
=
⇒
= 2 2
2
2
− b (a + b) − a (a + b ) a b (a + b)
1
1
⇒
u = 2 ,v = 2 .
a
b
Hence, x = a2, y = b2 is the solution of the given equations.
Q.6
Solve: x + y = a + b
ax – by = a2 – b2
Sol.
The given system of equations may be written as
x + y – (a + b) = 0
ax – by – (a2 – b2) = 0
By cross-multiplication, we get
x
−y
1
=
=
2
2
2
2
− a − b − (− b) × −(a + b ) − a − b − a × −(a + b ) 1× −b − a × 1
x
−y
1
=
=
⇒
2
2
2
2
2
2
− a + b − ab − b
− a + b + a + ab − b − a
x
−y
1
=
=
⇒
− a(a + b ) b(a + b ) − (a + b )
x
y
1
⇒
=
=
− a(a + b ) − b(a + b ) − (a + b )
− b(a + b )
− a(a + b)
x=
=b
⇒
= a and y =
− (a + b )
− (a + b)
Hence, the solution of the given system of equations is x = a, y = b.
x y
Q.7
Solve: + = 2
a b
ax – by = a2 – b2
Sol.
The given system of equations may be written as
bx + ay – 2ab = 0
ax – by – (a2 – b2) = 0
By cross-multiplication, we have
(
)
(
)
35
1
x
−y
=
=
2
2
− a a − b − (− b )(− 2ab) − b a − b − a(− 2ab) b × −b − a × a
−y
1
x
=
=
2
2
2
2
2
2
2
− a a − b − 2ab
− b a − b + 2a b − b − a 2
1
x
−y
=
=
2
2
2
2
2
2
2
− a a − b + 2b
− a − b − 2a
− a − b2
1
x
−y
=
=
2
2
2
2
2
− a +b
−b −a −b
− a + b2
(
⇒
⇒
⇒
2
2
(
(
)
(
(
)
(
)
2
(
)
)
(
)
(
)
(
)
− a(a + b )
− b(a + b )
x=
= a and y =
=b
− (a + b )
− (a + b )
2
⇒
)
2
2
)
2
2
2
2
Hence, the solution of the given system of equations is x = a, y = b.
36
Q.8
Solve the following system of equations in x and y
ax + by = 1
(a + b )2 − 1or, bx + ay = 2ab
bx + ay = 2
a + b2
a 2 + b2
Sol.
The given system of equations may be written as
ax + by = 1
2ab
bx + ay = 2 2
a +b
By cross-multiplication, we have
y
x
−1
=
= 2 2
2
2
2ab
2a b
a −b
−a b− 2 2
2
2
a +b
a +b
x
y
−1
= 2
= 2 2
⇒
2
3
2
3
2
2ab − a − ab
a b − b − 2a b a − b
2
2
a +b
a 2 + b2
−1
x
y
= 2 2
=
⇒
2
3
2
3
ab − a
−a b−b
a −b
2
2
2
2
a +b
a +b
3
2
a − ab
1
a 2b − b3
1
x= 2 2 × 2 2 ,y= 2 2 × 2 2
⇒
a +b
a −b
a +b
a −b
2
2
2
2
a a −b
1
b a −b
1
x= 2 2 × 2 2,y= 2 2 × 2 2
⇒
a +b
a −b
a +b
a −b
a
b
y= 2 2
x= 2 2 ,
⇒
a +b
a +b
(
)
(
)
Hence, the solution of the given system of equation is x =
a
b
,y= 2 2
2
a +b
a +b
2
Solve: a(x + y) + b (x - y) = a2 – ab + b2
……(1)
a(x + y) – b (x – y) = a2 + ab + b2
……(2)
Sol.
Taking x + y = u and x – y = v the given system of equations becomes
au + bv = (a2 – ab + b2)
……(3)
au – bv = (a2 + ab + b2)
…..(4)
Adding (3) and (4)
a 2 + b2
⇒ u=
2au = 2a2 + 2b2
a
a2 + b2
⇒
………(5)
x+ y=
a
Subtracting (4) from (3)
⇒ v = -a
2bv = -2ab
⇒ x – y = -aa ……(6)
Adding (5) and (6)
Q.9
37
a 2 + b2 a
−
2x =
a
1
2
2
a + b − a2
2x =
a
2
b
⇒
x=
2a
Subtracting (6) from (5)
a2 + b2
2y =
+a
a
a2 + b2 + a 2
2y =
a
2
2a + b 2
y=
⇒
2a
⇒ 2y =
2a 2 + b 2
a
Hence,
ence, the solution of the given system of equations is x =
Q.10
Solve: ax + by = c
bx + ay = 1 + c
multiplication, we have
Sol.
By cross-multiplication,
x
y
-1
b
c
a
b
a
1+c
b
a
x
y
−1
= 2 2
=
b(1 + c ) − ac bc − a(1 + c ) a − b
x
y
−1
=
= 2 2
⇒
b + bc − ac bc − a − ac a + b
− (b + bc − ac)
− (bc − a − ac)
x=
,
y=
⇒
2
2
a −b
a 2 − b2
− b − bc + ac
− bc + a − ac
x=
⇒
,
y=
(a − b)(a + b)
a 2 − b2
−b
c(a − b)
a
c(a − b )
x= 2 2 +
⇒
, y= 2 2 + 2 2
a − b (a − b)(a + b )
a −b
a −b
−b
c
a
c
x= 2 2 +
y= 2 2 +
,
⇒
a −b a+b
a −b a +b
Hence, the solution of the given system of equations is
c
a
c
b
+ 2 2
x=
− 2 2,y=
a +b a −b
a +b a −b
Conditions for Solvability or Consistency
Let the system of equations
a1x + b1y + c1 = 0
b2
2a 2 + b 2
,y=
2a
2a
38
a2x + b2y + c2 = 0
(i)
is consistent with unique solution, if
a1 b1
≠
a2 b2
i.e., line are Intersecting.
(ii)
is consistent with infinitely many solutions, if
a1 b1 c1
= =
a2 b2 c2
i.e., line are coincident.
a b c
(iii)
is inconsistent, if 1 = 1 ≠ 1
a2 b2 c2
i.e., lines are parallel and non
non-coincident.
Illustrative examples
In each of the following systems of equations determine whether the system has a
Q.1
unique solution, no solution or infinitely many solutions. In case there is a unique
solution, find it.
(i)
2x + 3y = 7
(ii)
6x + 5y = 11
(iii)
-3x + 4y = 5
15
15
9
9 x + y = 21
x − 6y + = 0
6x + 5y = 11
2
2
2
Sol.(i) The given system of equations may be written as
2x + 3y =7
6x + 5y =11
Where,
a1 = 2, b1 = 3, c1 = 7 and a2 = 6, b2 = 5, c2 = 11
a1 2 1
b 3
= = and 1 =
We have,
a2 6 3
b2 5
a1 b1
≠
Clearly,
a2 b2
So, the given system of equations has a unique solution.
x
y
−1
⇒
⇒
⇒
3
7
2
5
11
6
3
5
−1
x
y
=
=
33 − 35 42 − 22 10 − 18
x
y −1
=
=
− 2 20 − 8
2 −1
− 20 5
x=
=
and y =
=
−8 4
−8 2
1
5
Hence, the given system of equations has a unique solution given by x = − , y = .
4
2
(ii)
The given system of equations may be written as
6x + 5y = 11
39
9x +
Where,
15
y = 21
2
a1 = 6, b1 = 5, c1 = 11, a2 = 9, b2 =
15
, c2 = 21
2
5
2
a1 6 2
b1
c1 11
and
= = ,
=
=
=
a2 9 3
b2 15 / 2 3
c2 21
a1 b1 c1
= ≠
Clearly,
a2 b2 c2
in-consistent.
So, the given system of equations has no solution i.e. it is in
(iii)
9
15
Here, a1 = -3, b1 = 4, c1 = 5 and a2 = , b2 = −6, c2 = −
2
2
c1
−5 −2
a1 − 3 − 2 b1
4 −2
and =
=
=
=
=
, =
We have,
c2 15 / 2 3
a2 9 / 2 3 b2 − 6 3
a1 b1 c1
= =
Clearly,
a2 b2 c2
So, the given system of equations has infinitely many solutions.
Q.2
For each of the following system of equations determine the value of k for which the
given system of equations has a unique solution:
(i)
2x – 3y = 1
(ii)
2x + 3y – 5 = 0
kx + 5y = 7
kx – 6y – 8 = 0
Sol.(i) The given system of equations is
2x – 3y =1
kx + 5y = 7
where, a1 = 2, b1 = -3, c1 = 1 and a2 = k, b2 = 5, c2 = 7
For a unique solution,
a1 b1
2 −3
− 10
≠ i.e., ≠
⇒k ≠
a2 b2
k
5
3
So, the given system of equations is consistent with a unique solution for all values of k ot
other
than -10/3.
(ii)
The given system of equations is
2x + 3y = 5
kx – 6y = 8
It is of the form
a1x + b1y = c1
and,
a2x + b2y = c2
where, a1 = 2, b1 = 3, c1 = 5 and a2 = k, b2 = -6 and c2 = 8
For a unique solution, we must have
2 3
a1 b1
≠ i.e., ≠
⇒ k ≠ −4
a2 b2
k −6
We have,
40
So, the given system of equations will have a unique solution for all real values of k other than
-4.
41
Q.3
For what value of k will the pair of equations
 x + 2y = 3

5x + ky = −7
has (i) a unique solution (ii) no solution.
Is there any value of k for whic
which
h the pair has an infinite number of solutions? If yes,
what is its value?
Sol.
Here,
a1 = 1, b1 = 2, c1 = 3.
Also,
a2 = 5, b2 = k, c2 = -7.
a b
(i)
For unique Sol. 1 ≠ 1
a2 b2
1 2
≠
⇒
5 k
⇒
k ≠ 10.
Thus, the given pair has a unique solution if k ≠ 10.
a b c
(ii)
For no solution 1 = 1 ≠ 1 .
a2 b2 c2
1 2 −3
= ≠
5 k
7
1 2
2 −3
= and ≠
⇒
k
5 k
7
− 14
, i.e., when k = 10.
k = 10 and k ≠
3
(iii)
For the given pair of equations to have an infinite number of solutions, we must have
1 2 −3
a1 b1 c1
= = , i.e., = =
5 k 7
a2 b2 c2
1 −3
≠
Since
5 7
∴ The system will never have infinites
Q.4
Determine the value of a for which the following pair of linear equations has infinitely
many solutions.
ax + 3 y = a − 3

12x + ay = a
Sol.
Here, a1 = a, b1 = 3 and c1 = a-3. Also, a2 = 12, b2 = a and c2 = a
For infinitely many solutions
a1 b1 c1
= =
a2 b2 c2
a 3 a −3
= =
⇒
12 a
a
42
a 3
3 a −3
=
=
and
12 a
a
a
⇒
a2 = 36
and a2 -3a= 3a
⇒
a = ±6
and a2 – 6a = 0
⇒
a = ±6
and a(a – 6) =0
⇒
a = ±6
and
a = 0 or a = 6.
Therefore, the common solution is a = 6.
Thus, for a = 6, the given pair of equations has infinitely many solution.
Q.5
For what value of k will the equations
x + 2y + (11 – k) = 0 and 2x + ky + (10 + k) = 0
represent the coincident lines?
Sol.
Here, a1 = 1, b1 = 2, c1 = 11 – k
and a2 = 2, b2 = k, c2 = 10 + k
The given equations will represent coincident lines, if they have infinitely many
solutions. The condition for this is.
a1 b1 c1
= =
a2 b2 c2
1 2 11 − k
= =
⇒
2 k 10 + k
1 2
= ⇒ k = 4.
Now,
2 k
1 11 − k
=
⇒ 10 + k = 2(11 − k )
Also,
2 10 + k
⇒
3k = 12 ⇒ k = 4.
∴ The given equations will represent coincident lines, if k = 4.
Q.6
Find the values of α and β for which the following system of equations has infinite
number solutions:
2x + 3y = 7
(α + β) x + (2α - β) = 3 (α + β + 1)
Sol.
The given system of equation will have infinite number of solutions, if
α + β 2α − β 3(α + β + 1)
=
=
2
3
7
α + β 2α − β
α + β 3(α + β + 1)
=
=
and
⇒
2
3
2
7
⇒
3(α + β) = 2 (2α - β ) and 7(α + β) = 6(α + β + 1)
Subtracting these, we get
(a – 5β) – (α + β) = 0 – 6
⇒
-6β = -6 ⇒ β = 1.
Also,
α = 5β = 5.
Hence, the given equations have infinite number of solutions if α = 5 and β = 1.
Q.7
For what value of p, will the following system of linear equations has no solution?
⇒
43
3x + y = 1 and (2p – 1) x + (p – 1) y = (p + 1)2
Sol.
Here a1 = 3, b1 = 1, c1 = 1
And a2 = 2p – 1, b2 = p – 1, c2 = (p + 1)2
Now Eqs. (1) and (2) have no solution, if
a1 b1 c1
= ≠
a2 b2 c2
i.e., the given equations have no solution, if
3
1
1
=
≠
2 p − 1 p − 1 ( p + 1)2
3
1
1
1
≠
=
and
i.e. if
p − 1 ( p + 1)2
2 p −1 p −1
3
1
=
⇒ 3( p − 1) = 2 p − 1
Now,
2 p −1 p −1
⇒
p=2
1
1
≠
.
For this value of p clearly,
p − 1 ( p + 1)2
Hence, the given system of equations has not solution, if p = 2.
Q.8
Find the values of p and q for which the following system of equations has infinite
number of solutions:
2x + 3y = 7
(p + q) x + (2p - q) y = 21
Sol.
The given system of equations will have infinite number of solutions, if
2
3
7
=
=
p + q 2 p − q 21
2
3
1
=
=
⇒
p + q 2p − q 3
2
1
3
1
= and
=
⇒
p+q 3
2p − q 3
⇒
p+q=6
…….(i)
⇒
2p – q = 9
…….(ii)
⇒
(p + q) + (2p – q) = 6 + 9
⇒
3p = 15
⇒
p = 15
Putting p = 5 in Eqn. (i) 5 + q = 6
⇒
q=1
Hence, the given system of equations will have infinitely many solutions, if p = 5 and q = 1.
Q.9
Find the values of α and β for which the following system of linear equations has infinite
number of solutions:
2x + 3y = 7
44
2αx + (α + β) y = 28
Has (i) a unique solution? (ii) no solution?
Sol.
The given system of equations will have infinite number of solutions, if
2
3
7
=
=
2α α + β 28
1
3
1
=
=
⇒
α α +β 4
1 1
3
1
= and
=
⇒
α 4
α +β 4
⇒
α = 4 and α + β = 12
⇒
α = 4 and β = 8
Hence, the given system of equations will have infinitely many solutions, if α = 4 and β = 8.
Q.10 Determine the values of m and n so that the following system of linear equations have
infinite number of solutions:
(2m – 1) x + 3y – 5 = 0
3x + (n – 1) y – 2 = 0
The given system of equations will have infinite number of solutions, if
Sol.
2m − 1
3
−5
=
=
3
n −1 − 2
2m − 1
3
5
=
=
⇒
3
n −1 2
2m − 1 5
3
5
= and
=
⇒
3
2
n −1 2
⇒
4m – 2 = 15 and 6 = 5n – 5
⇒
4m = 17 and 5n = 11
17
11
m = and n =
⇒
4
5
Hence, the given system of equations will have infinite number of solutions, if
17
11
m = and n = .
4
5
Q.11 Determine the value of k so that the following linear equations have no solution:
(3k + 1) x + 3y – 2 = 0
(k2 + 1) x + (k – 2) y – 5 = 0
Sol.
The given system of equations will have no solution, if
3k + 1
3
−2
=
≠
2
k +1 k − 2 − 5
3k + 1
3
3
2
=
and
≠
⇒
2
k +1 k − 2
k −2 5
3k + 1
3
=
Now,
k2 +1 k − 2
45
(3k + 1) (k – 2) = 3(k2 + 1)
3k2 – 5k – 2 = 3k2 + 3
-5k – 2 = 3
-5k = 5
k = -1
3
2
≠ for k = −1.
Clearly,
k −2 5
Hence, the given system of equations will have no solution for k = -1.
Problems on Linear Equations
The sum of two number is 58. The greater number exceeds twice the smaller number by
Q.1
1. Find the numbers.
Let the smaller number be x and the greater number be y.
Sol.
Sum of the two numbers is 58.
∴
x + y = 58
The greater number exceeds twice the smaller number by 1.
∴
y = 2x +1
Substituting 2x + 1 for y in Eq. (1), we get
x + 2x + 1 = 58
⇒
3x + 1 = 58
⇒
3x = 57
Now, substituting x = 19 in Eq. (2) we get
y = 2 × 19 + 1 = 38 + 1 = 39.
Hence, the greater number is 39 and the smaller number is 19.
Q.2
If 3 times the larger
rger of the two number is divided by the smaller one we get 5 as
quotient and 1 as remainder. Also, if 7 times the smaller number is divided by the larger
one, we get 4 as quotient and 4 as remainder. Find the numbers
Sol.
Let x and y be larger and smaller numbers respectively. By Division Algorithm,
Dividend = Divisor × Quotient + Remainder.
(i)
When 3x is divided by y, we get 5 as quotient and 1 as remainder.
3x = y × 5 + 1 ⇒ 3x – 5y – 1 = 0
…….(1)
(ii)
When 7y is divided by x, we get 4 as quotient and 4 as remainder.
7y = x × 4 + 4 ⇒ 4x – 7y + 4 = 0
…….(2)
Solving Eqs. (1) and (2) we get
x = 27 : y = 16.
Hence, the required numbers are 27 and 16.
Q.3
The sum of the numerator and denominator of a fraction is 12. If the denominator is
1
increased by 3, the fraction becomes  . Find the fraction.
2
Sol.
Let the numerator and denominator of the fraction be x and y respectively. Then
x
fraction =
y
⇒
⇒
⇒
⇒
⇒
46
It is given
Numerator + denominator = 12
⇒
x + y = 12
Q.4
Sol.
Q.5
Sol.
…….(1)
1
Also, when denominator is increased by 3, the new fraction =
2
x
1
=
∴
y+3 2
⇒
2x = y + 3
⇒
2x – y = 3
……(2)
Now, Eq. (1) + Eq. (2)
⇒
3x = 15 ⇒ x = 5
From Eq. (1), y = 12 – x = 12 – 5 = 7.
5
Hence, the required fraction is
7
The sum of the present ages of Suresh and his mother is 89 years. After 11 years,
mother’s age will be twice Suresh’s age. Find ttheir present ages.
Let the present age of Suresh be x years and that of his mother be y ears.
The sum of their present ages is 89 years.
∴
x + y = 89
……(1)
After 11 years, mother’s age = (y + 11) years and Suresh’s age = (x + 11) years.
But after 11 years, mother’s age is twice that of Suresh’s age,
∴
y + 11 = 2(x + 11)
⇒
y + 11 = 2x + 22
⇒
y = 2x + 11
…….(2)
Substituting y=2x + 11 in Eq. (1), we get
x + 2x + 11 = 89
⇒
3x + 11 = 89
⇒
3x= 89 – 11 = 78
⇒
x = 26
Substituting the value x = 26 in Eq. (2), we get
y = 2 × 26 + 11 = 52 + 11 = 63.
Thus, Suresh’s age is 26 years and his mother’s age is 63 years.
If twice the son’s age in years is added to the father’s age, the sum is 70. But if twice the
father’s age is added to the son’s age
age,, the sum is 95. Find the ages of father and son.
Suppose father’s age (in years) be x and that of son’s be y. Then,
x + 2y = 70
…….(1)
2x + y = 95
…….(2)
Multiply Eq. (1) by 2 and Eqn. (2) by 1
2x + 4y = 140
……(3)
2x +y = 95
……(4)
Subtracting Eqn. (4) from (3)
3y = 45 ⇒ y = 15
47
Putting y = 15 in Eqn (1)
x + 2 × 15 = 70 ⇒ x = 40
∴ Age of father = 40 years
Age of son = 15 years.
48
Q.6
I am three times as old as my son. Five years later, I shall be two and a half times as old
as my son. How old am I and how old is my son?
Sol.
Suppose my age is x years and my son’s age is y years. Then,
x = 3y
…..(1)
Five years later, my age= (x + 5) ye
years
my son’s= (y + 5) years.
5
∴
x + 5 = ( y + 5)
2
2x+10=5y+25
⇒
2x – 5y – 15 = 0
……(2)
Putting x = 3y in equation (2),
), we get
6y – 5y – 15 = 0 ⇒ y = 15
Putting y = 15 in equation (i), we get
x = 45
Hence, my present age is 45 years and my son’s present age is 15 years.
Q.7
A and B are friends and their ages differ by 2 years. A’s father D is twice as old as A and
B is twice as old as his sister C. The age of D and C differ by 40 years. Find the ages of A
and B.
Let the ages of A and B be x and y years respectively. Then,
Sol.
x – y = ±2
y
years.
D’s age = 2x years. , C’s age =
2
Clearly, D is older than C
y
2 x − = 40 ⇒ 4 x − y = 80
∴
2
Thus, we have the following two systems of linear equations
x–y=2
…….(i)
and 4x – y = 80
…….(ii)
x – y = -2
……(iii)
and, 4x – y = 80
…….(iv)
Subtracting equation (i) from
rom equation (ii), we get
3x = 78 ⇒ x = 26
Putting x = 26 in equation (i), we get y = 24
Subtracting equation (iv) from equation (iii), we get
82
1
− 3x = −82 ⇒ x = = 27
3
3
82
Putting x =
in equation (iii), we get
3
82
88
1
y = + 2 = = 29
3
3
3
Hence, A’s age = 26 years and B’s age = 24 years
49
Or,
1
1
years and B’s age = 29 years.
3
3
Q.8
Ten years ago, father was twelve times as old as his son and ten years hence, he will be
twice as old as his son will be. Find their present ages.
Sol.
Let the present ages of father and son be x years and y years respectively.
Ten years ago, Father’s age = (x – 10) years
Son’s age = (y – 10) years
∴
x – 10 = 12 (y – 10) ⇒ x – 12 y + 110 = 0
Ten years later, Father’s age = (x + 110) years.
Son’s age = (y + 10)
∴
x + 10 = 2 (y + 10) ⇒ x+10=2y+20
⇒
x – 2y – 10 = 0
Subtracting (ii) from (i), we get
-10 y + 120 = 0 ⇒ 10y = 120 ⇒ y = 12
Putting y = 12 in (i), we get
x – 144 + 110 = 0 ⇒ x = 34
Thus, present age of father is 34 years and the present age of son is 12 years.
Q.9
Five years hence, father’s age will be three times the age of his son. Five years ago,
father was seven times as old as his son. Find their present ages.
Sol.
Let the present age of father be x years and the present age of son be y years.
Five years hence, Father’s age = (x + 5) years
Son’s age = (y + 5) years
Using the given information, we have
x + 5 = 3 (y + 5) ⇒ x – 3y – 10 = 0
…..…(i)
Five years ago, Father’s age = (x – 5) years
Son’s age = (y – 5) years
Using the given information, we get
(x – 5) = 7 (y – 5)
⇒
x-5=7y-35
⇒ x – 7y + 30 = 0
……..(ii)
Subtracting equation (ii) from equation (i), we get
4y – 40 = 0 ⇒ y = 10
Putting y = 10 in equation (i), we get
x – 30 – 10 = 0 ⇒ x = 40
Hence, present age of father is 40 years and present age of son is 10 years.
ASSIGNMENT
A father is three times as old as his son. After twelve years, his age will be twice as that
1.
of his son then. Find their present ages.
2.
Ten years later, A will be twice as old as B and five years ago, A was three times as old as
B, What are the present ages of A and B?
3.
Six years hence a man’s age will be three times the age of his son and three years ago he
was nine timess as old as his son was then. Find their present ages.
A’s age = 27
50
4.
The present
sent age of a father is three years more than three times the age of the son.
Three years hence father’s age will be 10 years more than twice the age of the son.
Determine their present ages.
5.
Two years ago, a father was five times as old as his son. Two years later, his age will be 8
more than three times the age of the son. Find the present ages of father and son.
6.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old
as Sonu. How old are Nuri and Sonu.
ASSIGNMENT ANSWERS
Father’s age = 36 years, Son’s age = 12 years
1.
2.
A’s present age = 50 years, B’s present age = 20 years.
3.
Man’s age = 30 years, Son’s age = 6 years
4.
Father’s age = 33 years, Son’s age = 10 years
5.
Father’s age = 42 years, Son’s age = 10 yea
years
6.
Nuri’s age = 50 years, Sonu’s age = 20 years
Q.1
A man sold
d a chair and a ttable
able for Rs. 850, thereby making a loss of 10% on the table and
a gain of 10% on the chair. By selling them together for Rs. 950, he would have made a
profit of 10% on the table and a loss of 10% on the chair. Find the cost price of each.
Sol.
Let the cost price of a chair be Rs. x and that of a table be Rs. y.
Using the first condition, we have
110x 90 y
⇒ 110x + 90 y = 85000
+
= 850
……(1)
100 100
Using the second condition, we have
90x 110 y
+
= 950
⇒ 90x + 110 = 95000
100 100
Eq. (1) and Eq. (2), can be rewritten as
11x + 9y = 8500
9x + 11y = 9500
Solving these equations, we get
X = 200, y = 700.
Thus, the cost price of a chair is Rs. 200 and that of a table is Rs. 700.
Q.2
The perimeter of a rectangle is 44 cm. Its length exceeds twice its breadth by 4 cm. Find
the length and the breadth of the rectangle.
Sol.
Let the length and the breadth of the rectangle (in cm) be x and y respectively.
∴ Perimeter of the rectangle ==2(x + y).
But the given perimeter is 44 cm.
∴
2(x + y) = 44
⇒
x + y = 22
The second condition gives
x = 2y + 4
Substituting the value of x from Eq. (2) in Eq. (1), we have
2y + 4 + y = 22
⇒
3y + 4 = 22
51
⇒
∴
3y = 22 – 4 = 18
y=6
52
Q.3
Sol.
Now, substituting y = 6 in Eq. (2), we have
x = 2 × 6 + 4 = 16
x = 16.
Thus, the length of the rectangle is 16 cm and its breadth is 6 cm.
Out of a 940 km journey, a part of the journey was covered by a morocar at a speed of
72 km/hr. The remaining part of the journey was covered by a train at a speed of 84
km/hr. If the total distance was covered in 12 hours, find the distance travelled by the
train and the time taken by it to cover that distance.
Let the time taken by the motorcar be x hours, and that by the train be y hours.
Using the second condition,
dition, we have
x + y = 12
……(1)
Using the first condition, we have
72x + 84y = 940
…….(2)
Multiplying Eq. (1) by 72 and subtracting it from Eq. (2), we have
72x + 84y = 940
72x + 72y = 864
(-)
(-)
(-)
12 y = 76
76 19
1
= = 6 = 6 hours 20 minutes
12 3
3
19
and Distance = 84 y = 84 × = 532km
3
Thus, the distance covered by the train is 532 km is 6 hours 20 minutes.
1
Suhel travelled 300 km by train and 200 km by taxi. It took him 5 hours. If he travels
2
260 km by train and 240 km by taxi, it takes 6 minutes longer. Find the speeds of the
train and taxi.
Let x km/hr and y km/hr be respectively the speeds of the train and taxi.
300
hours.
(i)
Time taken to cover 300 km by train =
x
200
Time taken to cover 200 km by taxi =
hours.
y
300 200 11
+
=
∴
……(1)
x
y
2
260
(ii)
Time taken to cover 260 km by train =
hours.
x
240
Time taken to cover 240 km by taxi =
hours.
y
Total time taken
en in this case = 5 hours 30 minutes + 6 minutes
y=
Q.4
Sol.
53
= 5 hours 36 minutes = 5
260 240 28
= .
+
…..(2)
5
x
y
1
1
= u and = v in Eq. (1) and Eq. (2), we get
Let
x
y
11
300u + 200v = ⇒ 600u + 400v = 11
……(3)
2
28
260u + 240v =
⇒ 1300u + 1200v = 28
……(4)
5
Solving Eqn (3) and (4)
1
1
u=
;v =
100
80
∴ x = 100 km/ hr, y = 80 km/hr.
These are the required speeds of the train and taxi respectively.
A two-digit
digit number is four times the sum of its digits and twice the product of the digits.
Find the number.
Let x be at the units place and y be at the tens place of the two
two-digit
git number. Then,
The number =10y + x.
According to the question,
10y + x = 4(x + y)
……(1)
and 10y + x = 2 × x × y
……(2)
Now, Eq. (1) ⇒ x = 2y Putting x=2y in eqn. (2)
From Eq. (2), 10y + 2y = 2 × (2y) × y
⇒
12y = 4y2
[Q y ≠ 0]
⇒
y=3
∴
x=2×3=6
Hence, the required number is
10y + x = 10 × 3 + 6 = 36
The sum of the digits of a two
two-digit
digit number is 12. The number obtained by reversing the
order of the digits of the given number exceeds the given number by 18. Find the given
number.
Let the tens digit of the given number be x and its units digit be y. Then
x + y = 12
……(1)
The given number = 10x + y.
The number obtained by reversing tthe order of the digits = 10y + x.
By the second condition, we have
(10y + x) – (10x + y) = 18
⇒
9y – 9x = 18
⇒
y–x=2
………(2)
Adding Eq. (1) and Eq. (2), we get
∴
Q.5
Sol.
Q.6
Sol.
3
hours.
5
54
Q.7
Sol.
Q.8
Sol.
2y = 14 ⇒ y = 7
Subtracting Eq. (2)
2) from Eq. (1), we get
2x = 10 ⇒ x = 5.
Therefore, the given number = 10 × 5 + 7 = 57.
The sum of a two-digit
digit number and the number obtained on reversing the order of its
digits is 132. If the digits differ by 2, find the number
Let x be the digit at unit’s place and y be the digit at ten’s place
Then, the number is 10y + x.
……(1)
Number obtained on reversing the order of the digits = 10x + y
Acc. To question (10y + x) + (10x + y) = 132 and x – y = 2 or y – x = 2
⇒
11x + 11y = 132
and
x – y = 2 or y – x = 2
⇒
x + y = 12
and
x – y = 2 or y – x = 2.
Thus, we obtain the following system of linear equations:
(i)
x + y = 12
or (ii) x + y = 12
x–y=2
y–x=2
Their solutions are respectively
(i)
x = 7, y= 5
or (ii) y = 7, x = 5
∴ From Eq. (1), we have
Number = 10 × 5 + 7 = 57 or 10 × 7 + 5 = 75
Hence, the required numbers are 57 or 75.
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go
40 km upstream and 55 km downstream. Determine the speed of the stream and of the
boat in still water.
Let thee speed of the boat in still water be x km/hr and the speed of the stream be y
km/hr.
Speed of the boat going upstream = (x – y) km/hr
Speed of the boat going downstream = (x + y) km/hr
dis tan ce
We know that, speed =
time
30
∴ Time taken by the boat in going 30 km upstream =
hours.
x− y
44
Time taken by the boat in going 44 km downstream =
hours.
x+ y
44
30
= 10
∴
+
x− y x+ y
Again,
40
Time taken by the boat in going 40 km upstream =
hours.
x− y
55
Time taken by the boat in going 55 km downstream =
55
hours.
x+ y
40
55
+
= 13
……(2)
x− y x+ y
1
1
= u and
= v in Eq. (1) and Eq. (2), we get
Substituting
x− y
x+ y
30u + 44v = 10
……..(3)
40u + 55v = 13
…….(4)
Solving these Eqn.
1
1
u = and v =
5
11
1
1
1
1
= and
=
⇒
x− y 5
x + y 11
⇒
x – y = 5 and x + y = 11.
On solving these two equations, we get x = 8 and y = 3.
Thus, the speed of the boat in still water is 8 km/hr and speed of the stream is 3 km/hr.
Q.9
In α ∆ ABC, ∠A = xo, ∠B = (3x – 2)o and ∠C = y°. Also, ∠C - ∠B = 9°. Determine the three
angles.
Sol.
We know that
∠A + ∠B + ∠C = 180°
∴
x + (3x – 2) + y = 180
⇒
4x + y – 2 = 180
⇒
4x + y = 182
182………(1)
Also, ∠C - ∠B = 9°
∴ y – (3x – 2) = 9
⇒ y – 3x + 2 = 9
⇒ y – 3x = 7 ………..(2)
Subtracting Eq. (2) from Eq. (1), we have
4x + y = 182
-3x + y = 7
(+) (-) (--)
7 x = 175
∴
∴
175
= 25
7
Substituting x = 25 in Eq. (1), we have
4 × 25 + y = 182
⇒
100 + y = 182
⇒
y = 182 – 100 = 82
∴
y = 82.
Hence, the three angles of ∆ ABC are
x=
56
Q.10
Sol.
Q.11
Sol.
Q.12
Sol.
∠A = x° = 25°, ∠B = (3x – 2)° = (75 – 2)° = 73° and ∠C = y° = 82°.
Anup went to a bank to withdraw Rs. 2000. He asked the cashier to give him Rs. 50 and
Rs. 100 notes only. Anup got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100
were received by him?
Let Anup received x notes of Rs. 50 and y notes of Rs. 100. Then,
x + y = 25
…….(1)
50x + 100y = 2000
…….(2)
From eq. (1), we have
y = 25 – x
Substituting y = 25 – x in Eq. (2), we have
50x + 100(25 – x) = 2000
⇒
50x + 2500 – 100x = 2000
⇒
50x = 500
⇒
x = 10
⇒
Y = 25 – x = 25 – 10 = 15
Thus, Anup received 10 notes of Rs. 50 and 15 notes of Rs. 100.
A part of the monthly hostel charges in a college are fixed and the remaining depend on
the number of days one has taken food in the mess. Student A takes food for 20 days
and pays Rs. 1000 as hostel charges, whereas student B takes food for 26 days and pays
Rs. 1180 as hostel charges. Find the fixed charged and the cost of food per day.
Let x be the number of days a student eats in the mess and y be the fixed monthly
charges in the hostel.. Then,
20x + y=1000;
26x + y=1180 .
Solving these, we get
x = 30 and y = 400.
∴
Fixed charges = Rs. 400
Cost of food per days = Rs. 30.
Points A and B are 90 km apart from each other on a highway. A car starts from A and
another from B at the same time. If they go in the same direction they meet in 9 hours
and if they go in opposite directions they mee
meett in 9/7 hours. Find their speeds.
Let speed of car at A = x km/hr
Speed of car at B = y km/hr
Let x is more than y
When they move in same direction
Their Relative speed = (x – y) km/hr
Time = 9 hrs.
Distance = 90 km
Speed × Time = 90 km
(x – y) 9 = 90
⇒
x – y = 10
…..(1)
When they move in Opp. Direction
Their relative speed = (x + y) km/hr
57
9
hrs.
7
Distance = 90 km
9
⇒
(x + y)
= 90
7
x + y = 70
……(2)
Solving (1) and (2) x = 40, y = 30
Q.13 A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km
upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water
and that of the stream.
Let the speed of the boat in still water be x km/hr and the speed of the stream be y
Sol.
km/hr. Then,
Speed in upstream = (x – y) km/hr
Speed in downstream = (x + y) km/hr
32
Now, Time taken to cover 32 km upstream =
hrs
x− y
36
Time taken to cover 36 km downstream =
hrs
x+ y
But, total time of journey
rney is 7 hours.
32
36
+
=7
∴
…….(i)
x− y x+ y
40
Time taken to cover 40 km upstream =
x− y
48
Time taken to cover 48 km downstream =
x+ y
In this case, total time of journey is given to be 9 hours.
40
48
+
=9
∴
……...(ii)
x− y x+ y
1
1
= u and
= v in equations (i) and (ii), we get
Putting
x− y
x+ y
32u + 36v = 7
……..(iii)
40u + 48v = 9
……..(iv)
Solving these equations, we get
1
1
u = and v =
⇒
8
12
1
1
1
= ⇒ x− y =8
Now, u = ⇒
x− y 8
8
1
1
1
= ⇒ x + y = 12
and, v = ⇒
x + y 12
12
Time =
58
Solving equations we get x = 10 and y = 2.
Hence, Speed of the boat in still water = 10 km/hr
Speed of the stream = 2 km/hr.
59
ASSIGNMENT
Points A and B are 70 km a part on a highway. A car starts from A and another car starts
1.
from B simultaneously. If they tra
travel
vel in the same direction, they meet in 7 hours, but if
they travel towards each other, they meet in one hour. Find the speed of the two carts.
2.
A boat goes 30 km upstream and 44km downstream in 10 hours. In 13 hours, it can go
40 km upstream and 55 km d
downstream.
ownstream. Determine the speed of stream and that of the
boat in still water.
3.
While covering a distance of 30 km. Ajeet takes 2 hours more than Amit. If Ajeet doubles
his speed, he would take 1 hour less than Amit. Find their speeds of walking.
4.
A man walks a certain distance with certain speed. If he walks 1/2 km an hour faster, he
takes 1 hour less. But, if he walks 1 km an hour slower, he takes 3 more hours. Find the
distance covered by the man and his original rate of walking.
5.
Places A and B are 80 km apart from each other on a highway. A car starts from A and
other from B at the same time. If they move in the same direction, they meet in 8 hours
and if they move in opposite directions, they meet in 1 hour and 20 minutes. Find the
speeds of the cars.
A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km
6.
upstream and 32 km downstream in the same time. Find the speed of the boat in still
water and the speed of the stream.
7.
Ritu can row downstream 20 km in 2 hours, and upst
upstream
ream 4 km in 2hours. Find her
speed of rowing in still water and the speed of the current.
8.
Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But
if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the
speed of the train and that of the taxi.
9.
Places A and B are 100 km apart on the highway. One car starts from A and another
from B at the same time. If the cars travel in the same direction at different speeds, they
meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the
speeds of two cars?
ASSIGNMENT ANSWERS
1.
Speed of car starting from point A = 40 km/hr.
Speed of car starting from point B = 30 km/hr.
2.
Speed of stream = 3km/hr, Speed of boat = 8km/hr.
3.
Ajit’s speed = 5km/hr, Amit’s speed = 7.5 km/hr.
4.
Distance – 36 km, original speed = 4 km/hr.
5.
35 km/hr, 25km/hr.
6.
6km/hr, 2km/hr.
7.
6 km/hr, 4km/hr.
8.
100 km/hr, 80 km/hr.
9.
60 km/hr, 40 km/hr.
60
ILLUSTRATIVE EXAMPLES
The taxi charges in a city comprise of a fixed charge for the distance covered. For a
Q.1
journey of 10 km the charge paid is Rs 75 and for a journey of 15 km the charge paid is
Rs. 110. What will a person have to pay for travelling a distance of 25 km?
Sol.
Let the fixed charges of taxi be Rs. x per km and the running charges be Rs. y km/hr.
According to the given condition, we have
x + 10y = 75
……(i)
x + 15y = 110
…….(ii)
Subtracting equation (ii) from equation (i), we get
-5 y= -35 ⇒ y = 7
Putting y = 7 in equation (i), we get x = 5.
∴
Total charges from travelling a distance of 25 km.
= x + 25y
= Rs. (5 + 25 × 7) = Rs. 180.
Q.2
The total expenditure per month of a household consists of a fixed rent of the house
and mess charges depending upon the number of people sharing the house. The total
monthly expenditure is Rs. 3900 for 2 people and Rs. 7500 for 5 people. Find the rent of
the house and the mess charges per head per month.
Sol.
Let the monthly rent of the house be Rs. x and the mess charges per head per month be
Rs. y.
According to the given conditions, we have,
x + 2y = 3900
……(i)
x + 5y = 7500
……(ii)
Subtracting equation (ii) from equation (i), we get
-3y = -3600 ⇒ y = 1200
Putting y = 1200 in equation (i), we get x = 1500
Hence, monthly rent = Rs 1500 and mess charges per head per month = Rs 1200.
Q.3
A person invested some amount at the rate of 12% ssimple
imple interest and some other
amount at the rate of 10% simple interest. He received yearly interest of Rs 130. But if
he had interchanged the amounts invested, he would have received Rs 4 more as
interest. How much amount did he invest at different rates?
Sol.
Suppose the person invested Rs x at the rate of 12% simple interest and Rs y at the rate
of 10% simple interest. Then,
12 x 10 y
+
Yearly interest =
100 100
12 x 10 y
+
= 130
∴
100 100
⇒
12x + 10y = 13000
⇒
6x + 5y = 6500
……(i)
If the invested amounts
ounts are interchanged, then yearly interest increases by Rs. 4.
10 x 12 y
+
= 134
∴
100 100
61
⇒
10x + 12y = 13400
⇒
5x + 6y = 6700
…….(ii)
Subtracting equation (ii) from equation (i), we get
x – y = -200
…….(iii)
Adding equation (ii) and (i), we get
11x + 11y = 13200
⇒
x + y = 1200
…….(iv)
Adding equation (iii) and (iv), we get
2x = 1000 ⇒ x = 500
Putting x = 500 in equation (iii), we get y = 700
Thus, the person invested Rs 500 at the rate of 12% per year and Rs. 700 at the rate of 10% per
year.
piece
ce of work in 10 days while 6 men and 8 boys can
Q.4
8 men and 12 boys can finish a pie
finish it in 14 days. Find the time taken by one man alone and that by one boy alone to
finish the work.
Suppose that one man alone can finish the work in x days and one boy alone can finish it
Sol.
in y days. Then,
1
One man’s one day’s work =
x
1
One boy’s one day’s work =
y
8
∴
Eight men’s one day’s work =
x
12
12 boy’s one day’s work =
y
Since 8 men and 12 boys can finish the work in 10 days
 8 12 
80 120
=1
10 +  = 1 ⇒
+
…..(i)
y
x
y
x


Again, 6 men and 8 boys can finish the work in 14 days.
6 8
84 112
14 +  = 1 ⇒
+
=1
∴
x
y
x y
……(ii)
1
1
= u and = v in equations (i) and (ii), we get
x
y
80u + 120v – 1 = 0
84u + 112v – 1 = 0
1
1
1
⇒ =
⇒ x = 140
Now solving these equations u =
140
x 140
1
1
1
⇒ =
⇒ y = 280.
And, v =
y 280
280
Putting
62
Thus, one man alone can finish the work in 140 days and one boy alone can finish the work in
280 days.
Q.5
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditure is 4 : 3. If
each of them saves Rs. 200 per month, find their monthly incomes.
Sol.
Let the income of first person be Rs. 9x and the income of second person be Rs. 7x.
Further,
er, let the expenditures of first and second person be 4y and 3y respectively. Then,
Saving of first person = 9x – 4y
Saving of second person = 7x – 3y
∴
9x – 4y = 200
……(i)
and, 7x – 3y = 200
……(ii)
Solving equation (i) and (ii) by
x = 200 and y = 400.
Thus, monthly income of first person = Rs 9x = Rs(9 × 200) = Rs 1800
and, monthly income of second person = Rs. 7x = Rs. (7 × 200) = Rs. 1400.
ASSIGNMENT
The area of a rectangle remains the same if the length is increased by 7 metres and the
1.
breadth in decreased by 3 metres. The area remains unaffected if the length is
decreased by 7 metres and breadth is increased by 5 metres. Find the dimensions of the
rectangle.
2.
A and B each has some money. If A gives Rs 30 to B, then B will have twice the money
left with A. But, if B gives Rs 10 to A, then A will have thrice as much as is left with B.
How much money does each have?
3.
There are two examination rooms A and B. If 10 candidates are sent from A to B, the
number of students in each room is same. If 220
0 candidates are sent from B to A, the
number of students in B is twice the students in A.. Find the number of students in each
room.
4.
In a ∆ ABC, ∠A = x°, ∠B = 3x° and ∠C = y°. If 3y – 5x = 30, prove that the triangle is right
angled.
5.
A part of monthlyy hostel charges in a college are fixed and the remaining depend on the
number of days one has taken food in the mess. When a student A takes food for 20
days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for
26 days, pays Rs 1180 as hostel charges. Find the fixed charge and the cost of food per
day.
2 Women and 5 men can together finish a piece of embroidery in 4 days, while 3 women
6.
and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the
embroidery,
idery, and that taken by 1 man alone.
7.
A wizard having powers of mystic in candations and magical medicines seeing a cock,
fight going on, spoke privately to both the owners of cocks. To one he said; if your bird
wins, than you give me your stake
stake-money, but if you do not win, I shall give you two
third of that’. Going to the other, he promised in the same way to give three fourths.
From both of them his gain would be only 12 gold coins. Find the stake of money each of
the cock-owners have.
63
8.
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark
for each wrong answer. Had 4 marks been awanded for each correct answer and 2
marks been deducted for each incorrect anser, then Yash would have scored 50 marks.
How many questions were there in the test?
9.
The students of a class are made to stand in rows. If 3 students are extra in a row, there
would be 1 row less. If 3 students are less in a row there would be 2 rows more. Find the
number of students in the class.
10.
One says,” give me hundred, friend! I shall then become twice as rich as you” The other
replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the
amount of their respective capital?
11.
ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y – 5)°, ∠C = (4x)° and
∠D = (7x + 5)°. Find the four angles.
ASSIGNMENT ANSWERS
1.
Length = 28m, Breadth = 15m
2.
A : Rs. 62, B : Rs. 34 3.
100, 80
5.
Fixed charge = Rs. 400; Cost of food per day = Rs. 30.
6.
36 days, 18 days
7.
42, 40 Gold Coins
8.
20
9.
36
10.
Rs. 40, Rs. 170
11.
∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°