Download Chapter 7 Linear Equations in Two Unknowns

7
Linear Equations in Two
Unknowns
Case Study
7.1 Linear Equations in Two Unknown and Their
Graphs
7.2 Solving Simultaneous Linear Equations in Two
Unknowns by the Graphical Method
7.3 Method of Substitution
7.4 Method of Elimination
7.5 Applications of Simultaneous Linear Equations
in Two Unknowns
Chapter Summary
Case Study
Can you find the actual number Hey, these chicks and
of chicks and rabbits there?
rabbits are very cute.
Suppose there are x chicks
and y rabbits.
According to the above
information, we can set up
2 equations as follows.
These animals have a
total number of 18 heads
and 46 legs.
 x  y  18 ............... (1)

 2x  4y  46 ........... (2)
From (1), we have x  18  y. ......... (3)
Substituting (3) into (2), we have:
2(18  y)  4y  46
Substituting y  5 into (3), we have
36  2y  4y  46
x  18  5
2y  10
 13
∴ There are 13 chicks and 5 rabbits.
y 5
P. 2
7.1 Linear Equations in Two Unknowns
and Their Graphs
In Book 1A, we have learnt about linear equations in one unknown.
In this chapter, we are going to learn about linear equations in two
unknowns.
Examples of linear equations in two unknowns:

x  y  18

3x  5y  7

2y  4x 
x
4
2
Each of the above equations consists of 2 unknowns with degree 1.
If one of the 2 variables is known, we can find the other by the
method of substitution.
P. 3
7.1 Linear Equations in Two Unknowns
and Their Graphs
A. Solutions of Linear Equations in Two Unknowns
Consider the equation x  y  10.
There are many pairs of values of x and y that satisfy this equation:
and y  10

and y  9

and y  8

and y  7
⋮
Each of the above pairs of x and y is a solution of the equation.
Besides the above 4 pairs, there are infinitely many solutions for the
linear equation in two unknowns.
We can write the solutions in ordered pairs as (0, 10), (1, 9), (2, 8), (3, 7),
etc., or by using a table:
x
0
1
2
3
…
y
10
9
8
7
…

x0
x1
x2
x3
P. 4
7.1 Linear Equations in Two Unknowns
and Their Graphs
A. Solutions of Linear Equations in Two Unknowns
Example 7.1T
Determine whether the following pairs are solutions of the equation
2x  y  4.
(a) (4, 0)
(b) (3, 2)
Solution:
(a) Substitute (4, 0) into the equation.
L.H.S.  2(4)  0
8
 R.H.S.
∴ (4, 0) is not a solution of the equation.
(b) Substitute (3, 2) into the equation.
L.H.S.  2(3)  (2)
4
 R.H.S.
∴ (3, 2) is a solution of the equation.
P. 5
7.1 Linear Equations in Two Unknowns
and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
Consider the equation y  x  2.
We write the solutions using a table:
x
y
2
0
1
1
0
2
1
3
…
…
For each pair of x and y, plot them onto the
rectangular coordinate plane and join them.
Consider some points from the line obtained such as (3, 1).
Check if this point satisfies the equation y  x  2:
L.H.S.  1
R.H.S.  3  2
 1
∵ L.H.S.  R.H.S.
∴ (3, 1) satisfies the equation y  x  2.
P. 6
7.1 Linear Equations in Two Unknowns
and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
In general, every point on the line satisfies
the equation y  x  2.
The straight line obtained is called the graph
of the equation y  x  2.
(1.8, 3.8)
(0.7, 2.7)
(0.4, 1.4)
(1.5, 0.5)
(3.5, 1.5)
Any point on the graph of a linear equation in two unknowns is
a solution of the equation.
On the other hand, if a point satisfies a linear equation in two
unknowns, the it lies on the graph of the equation.
P. 7
7.1 Linear Equations in Two Unknowns
and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
Example 7.2T
The figure shows the graph of 2x  4y  1.
Find the coordinates of C.
Solution:
The y-coordinate of C is 1.
Substitute y  1 into the equation 2x  4y  1, we have
2x  4(1)  1
2x  3
3
x 
2
 3

∴ The coordinates of C are   ,  1 .
 2

P. 8
7.1 Linear Equations in Two Unknowns
and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
Example 7.3T
(a) Draw the graph of the equation 2y  x  2. (Take x from 2 to 2.)
(b) Using the graph in (a), answer the following questions.
(i) If P(3, y) lies on the graph of 2y  x  2, find the value of y.
(ii) Is (3, 2.5) a solution of the equation 2y  x  2?
Solution:
(a)
x
y
(b) (i) ∵
∴
(ii) ∵
∴
2
0
2y  x  2
0
1
2
2
P(3, 0.5) lies on the
graph of 2y  x  2.
y  0.5
(3, 2.5)
(3, 0.5)
(3, 2.5) lies on the graph of 2y  x  2.
(3, 2.5) is a solution of the equation 2y  x  2.
P. 9
7.1 Linear Equations in Two Unknowns
and Their Graphs
B. Graphs of Linear Equations in Two Unknowns
The equation of a straight line can be expressed in different forms.
For example, y  4x  5 can be expressed as y  4x  5 or 4x  y  5  0.
However their graphs are the same.
Consider the graphs of equations in the form y  ax  b, where a and b
are constants.
(a) If a is fixed, consider:
(b) If b is fixed, consider:
(i) y  x  3,
(i) y  2x  1,
(ii) y  x and
(ii) y  x  1 and
(iii) y  x  3.
(iii) y  3x  1.
The graphs are parallel lines.
P. 10
The graphs pass through the same
point on the y-axis.
7.2 Solving Simultaneous Linear Equations in
Two Unknowns by the Graphical Method
The figure shows 2 linear equations y  x  1 and x  y  3.
The coordinates of A, C and other
points on the graph of y  x  1 satisfy
the equation y  x  1.
The coordinates of B, C and other
points on the graph of x  y  3 satisfy
the equation x  y  3.
We observe that C(1, 2) satisfies both
equations y  x  1 and x  y  3
simultaneously.
In fact, C(1, 2) is the only point (point of intersection) which lies on
both graphs.
So, (1, 2) is the common solution of the 2 equations.
P. 11
7.2 Solving Simultaneous Linear Equations in
Two Unknowns by the Graphical Method
We usually represent the pairs of equations as
yx1

,
xy3
which is called a pair of simultaneous
linear equations in two unknowns.
Solving a pair of simultaneous linear
equations by finding the point of
intersection of their graphs is called
the graphical method.
P. 12
7.2 Solving Simultaneous Linear Equations in
Two Unknowns by the Graphical Method
Example 7.4T
The figure shows the graph of the equation x  y  2.
(a) Complete the following table for the equation y  3x  4.
x
1
2
2.5
y
(b) Draw the graph of y  3x  4 in the same coordinate plane.
x y 2
(c) Hence solve 
graphically.
y

3
x

4

y  3x  4
Solution:
(a)
x
y
1
1
2
2
2.5
3.5
(b) Refer to the graph.
(c) From to the graph, the solution is x  1.5, y  0.5.
P. 13
(1.5, 0.5)
7.2 Solving Simultaneous Linear Equations in
Two Unknowns by the Graphical Method
Example 7.5T
y x5 0
Solve the simultaneous equations 
4 y  3x  8 graphically.

Solution:
yx50
x
y
5
0
3
–2
1
–4
–2
0.5
0
2
4y  3x  8
x
y
–4
–1
From to the graph, the solution is x  4, y  1.
P. 14
(4, 1)
7.2 Solving Simultaneous Linear Equations in
Two Unknowns by the Graphical Method
The figure shows the graphs of the linear equations 7y  9x  16
and 6y  5x  1  0.
From the graph,
x  1.2 and y  0.8.
We do not know the exact solution by
reading from the graphs.
(1.2, 0.8)
We observe that the drawing of straight
lines or the point of intersection of the
graphs may not be accurate.
The reading of the solution may depend on the scale of the grid lines.
Hence, the solutions obtained by the graphical method are
approximations only.
P. 15
7.3 Method of Substitution
In the last section, we have learnt how to solve simultaneous linear
equations in two unknowns by the graphical method.
However, as we know that there is a limitation to use this method, we
will explore other methods (algebraic methods).
There are 2 algebraic methods:
 Method of Substitution
 Method of Elimination
In this section, we are going to study the method of substitution.
P. 16
7.3 Method of Substitution
Consider the following simultaneous equations.
 x  3y  1 .................. (1)

 5x  6y  1 .............. (2)
The following shows the steps to apply the method of substitution.
Step 1: Make one of the unknowns, x or y, the subject of the equation.
Step 2: Substitute the result (3) into equation (2) to get a linear equation
in one unknown. Then solve this equation.
Step 3: Substitute the result (the value of y) obtained into one of the
above equation to find the remaining unknown.
From (1), we have x  1  3y ............ (3)
2
∴
x 13—
5(1  3y)  6y  1
3
5  15y  6y  1
 1
9y  6
2
2 .
y —
∴ The solution is x  1, y  —
3
3
P. 17
7.3 Method of Substitution
Example 7.6T
Solve the following simultaneous equations.
 y  x  1 .................. (1)

 4x  y  4 .............. (2)
Solution:
Substitute (1) into (2):
4x  (x  1)  4
4x  x  1  4
3x  3
x  1
Substitute x  1 into (1):
y  1  1
0
∴ The solution is x  1, y  0.
P. 18
7.3 Method of Substitution
Example 7.7T
Solve the simultaneous equations 2x  y  1  x  y  5.
Solution:
Rewriting the given equations, we have:  2x  y  1  5 ......... (1)
 x  y  5 ................. (2)
From (2), we have x  y  5 ............ (3)
Substitute (3) into (1):
2(y  5)  y  1  5
2y  10  y  1  5
3y  6
y  2
Substitute y  2 into (3):
x  2  5
3
∴ The solution is x  3, y  2.
P. 19
7.3 Method of Substitution
Example 7.8T
Solve the following simultaneous equations.
 4x  3y  10 ............ (1)

 6y  2x  8 ............ (2)
Solution:
From (2), we have 3y  x  4
x  3y  4 ............ (3)
Substitute (3) into (1):
4(3y  4)  3y  10
12y  16  3y  10
9y  6
2
y 
3
∴
2
The solution is x  2, y   .
3
P. 20
2
into (3):
3
 2
x  3    4
 3
2
Substitute y  
7.4 Method of Elimination
Besides the method of substitution, we can use the method of
elimination to solve simultaneous linear equations in two unknowns.
The key idea of this method is to eliminate one of the unknowns by
adding or subtracting the simultaneous equations.
Consider the following simultaneous equations.
 x  y  3 .............. (1)

 x  y  1 .............. (2)
(1)  (2):
OR
x y 3
) x  y  1
2x
4
x
2
Substitute x  2 into (1), we have y  1.
∴ The solution is x  2, y  1.
P. 21
(x  y)  (x  y)
xyxy
2x
x
31
4
4
2
7.4 Method of Elimination
Example 7.9T
Solve the following simultaneous equations by the method of elimination.
 4x  y  14 ............ (1)

 4x  5y  2 ............ (2)
Solution:
(1)  (2):
(4x  y)  (4x  5y)  14  (2)
4x  y  4x  5y  14  2
6y  12
y  2
Substitute y  2 into (1):
4x  (2)  14
4x  12
x  3
∴ The solution is x  3, y  2.
P. 22
OR
4x  y
) 4x  5y
6y
y
 14
 2
 12
 2
7.4 Method of Elimination
If the coefficients of x (or y) terms in a pair of simultaneous equations
are different, we should multiply the equations by some constants so
that we can eliminate those terms by addition or subtraction.
For example, consider the following simultaneous equations:
 x  2y  6 .............. (1)

 3x  y  11 ............ (2)
We can multiply equation (1) by 3 so that the coefficients of x of the
pair of simultaneous equations are the same:
(1)  3,
3(x  2y)  6  3
3x  6y  18 ......... (3)
Hence the coefficients of x in equations (2) and (3) are the same.
P. 23
7.4 Method of Elimination
Example 7.10T
Solve the following simultaneous equations by the method of elimination.
 3y  2x  47 ............ (1)

 4y  x  48 .............. (2)
Solution:
(2)  2: 8y  2x  96 ...... (3)
(1)  (3):
(3y  2x)  (8y  2x)  47  96
3y  2x  8y  2x  143
11y  143
y  13
Substitute y  13 into (2):
4(13)  x  48
x 4
∴ The solution is x  4, y  13.
P. 24
7.4 Method of Elimination
Example 7.11T
Solve the following simultaneous equations by the method of elimination.
 3x  2y  2 .............. (1)

 4x  5y  28 .......... (2)
Solution:
The L.C.M. of 3 and 4 is 12.
(1)  4: 12x  8y  8 ......... (3)
(2)  3: 12x  15y  84 ..... (4)
Eliminate x.
(3)  (4):
Suppose we choose to
eliminate x. Consider
(12x  8y)  (12x  15y)  8  (84)
the L.C.M. of 3 and 4.
23y  92
Then multiply some
constants on both sides
y 4
of the equations such
Substitute y  4 into (1):
that we can eliminate x.
3x  2(4)  2
x  2
∴ The solution is x  2, y  4.
P. 25
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
In our daily lives, we often come across problems that can be
formulated as simultaneous equations.
We can solve these problems by applying the technique of solving
simultaneous equations.
Step 1: Identify the 2 unknowns and represent them with letters such
as x and y.
Step 2: Set up a pair of simultaneous linear equations in 2 unknowns.
Step 3: Solve the simultaneous equations.
P. 26
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.12T
The difference between 2 numbers is 8. If the smaller number is doubled,
the result is 2 more than the larger number. Find the two numbers.
Solution:
Let x and y be the smaller and the larger numbers respectively.
From the problem, we have
 y  x  8 ............ (1)

 2x  y  2 .......... (2)
(1)  (2):
(y  x)  (2x  y)  8  2
x  10
Substitute x  10 into (1):
y  10  8
y  18
∴ The two numbers are 10 and 18.
P. 27
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.13T
Lily has some $3 and $5 stamps. If the total number of stamps is 20 and
their total value is $70, find the number of $3 and $5 stamps that she has.
Solution:
Let x and y be the numbers of $3 and $5 stamps respectively.
From the problem, we have:  x  y  20 .......... (1)
 3x  5y  70 ...... (2)
(1)  3: 3x  3y  60 ..... (3)
(2)  (3):
(3x  5y)  (3x  3y)  70  60
2y  10
y 5
Substitute y  5 into (1), we have x  5  20
x  15
∴ Lily has 15 $3 stamps and 5 $5 stamps.
P. 28
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.14T
A company produces washing powder in 2 kinds of packages X and Y.
The weight of one box of X is 50 g less than 2 boxes of Y. If the total
weight of 2 boxes of X and 1 box of Y is 2 kg, find the weight of each
package of washing powder.
Solution:
Let x g and y g be the weights of each box of X and Y respectively.
From the problem, we have:  2y  x  50 .......... (1)
 2x  y  2000 ...... (2)
(1)  2: 4y  2x  100 ..... (3)
(2)  (3):
Substitute y  420 into (2):
(2x  y)  (4y  2x)  2000  100
2x  420  2000
5y  2100
2x  1580
y  420
x  790
∴ The weights of each box of X and Y are 790 g and 420 g respectively.
P. 29
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.15T
Dickson is 3 years older than his sister Mary. 2 years ago, Dickson was
twice as old as Mary. How old is Dickson now?
Solution:
Let x and y be the present ages of Dickson and Mary respectively.
From the problem, we have:  x  y  3 ................. (1)
 x  2  2(y  2) ...... (2)
From (2), we have x  2  2y  4
x  2y  2 ........................... (3)
Substitute (3) into (1):
(2y  2)  y  3
y 5
Substitute y  5 into (1), we have x  5  3
x 8
∴ Dickson is 8 years old now.
P. 30
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.16T
3
Fanny finished a test with 2 parts A and B. She used of the time to
5
finish part A. If she used 20 minutes more to finish part A than part B,
(a) find the time she used to finish parts A and B respectively,
(b) find the time she used to finish the whole test.
Solution:
(a) Let x minutes and y minutes be the time she used to finish parts
A and B respectively.
 x  y  20 .......... (1)
From the problem, we have: 
3
 x  (x  y) ....... (2)
From (1), we have
5
Substitute (3) into (4):
x  y  20 .................. (3)
From (2), we have
2(y  20)  3y
5x  3(x  y)
y  40
2x  3y ...................... (4)
Substitute y  40 into (3),
x  60
∴ Part A : 60 minutes ; Part B : 40 minutes
P. 31
7.5 Applications of Simultaneous Linear
Equations in Two Unknowns
Example 7.16T
3
Fanny finished a test with 2 parts A and B. She used of the time to
5
finish part A. If she used 20 minutes more to finish part A than part B,
(a) find the time she used to finish parts A and B respectively,
(b) find the time she used to finish the whole test.
Solution:
(b) Part A : 60 minutes ; Part B : 40 minutes
The time she used to finish the whole test
 (60  40) minutes
 100 minutes
P. 32
Chapter Summary
7.1 Linear Equations in Two Unknowns and Their Graphs
1.
An equation consisting of 2 unknowns with degree 1 is called
a linear equation in two unknowns.
2.
Any linear equation in two unknowns has infinitely many
solutions which can be represented by ordered pairs.
3.
The graph of a linear equation in two unknowns is a straight
line.
4.
Any point on the graph of a linear equation in two unknowns
is a solution of the equation.
P. 33
Chapter Summary
7.2 Solving Simultaneous Linear Equations in Two
Unknowns by the Graphical Method
1.
The common solution of a pair of simultaneous equations
satisfies both equations.
2.
Solving simultaneous equations by the graphical method
involves finding the coordinates of the point of intersection
of the graphs of the 2 equations.
3.
Solutions obtained from solving simultaneous equations by
the graphical method are approximations only.
P. 34
Chapter Summary
7.3 Method of Substitution
The steps involved in solving simultaneous equations by the method
of substitution are as follows.
1.
Make one of the unknowns the subject of the equation.
2.
Substitute the result obtained into the other equation to get a
linear equation in one unknown. Then solve this equation.
3.
Substitute the result obtained into one of the given equations
to find the remaining unknown.
P. 35
Chapter Summary
7.4 Method of Elimination
1.
Solving simultaneous equations by the method of elimination
involves eliminating one of the unknowns by adding or
subtracting the simultaneous equations.
2.
If the coefficients of the x (or y) terms in a pair of simultaneous
equations are different, we should multiply one or both of the
equations by some constants so that we can eliminate those
terms by addition or subtraction.
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Chapter Summary
7.5 Applications of Simultaneous Linear Equations
in Two Unknowns
We can solve some daily-life problems by formulating simultaneous
equations and solving them through the following steps.
1.
Identify the 2 unknowns and represent them with letters.
2.
Set up a pair of simultaneous linear equations in 2 unknowns.
3.
Solve the simultaneous equations.
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