Download Systems of linear equations (simultaneous equations)

Before starting this topic you should review how to graph equations of lines. The link below will
take you to the appropriate location on the Academic Skills site.
http://www.scu.edu.au/academicskills/numeracy/index.php/12
The topics you need to review are listed below.
Topic 1: Examples of linear relationships
Topic 2: Graphing lines
Topic 3: Finding equations of lines
Systems of linear equations (simultaneous equations)
The example below will serve as an introduction to this module.
Example; If the ABC Taxi company charges a flag fall of $5 plus $1 per km, this can be modelled
by the linear equation: C= 1k + 5 where C is the cost to travel k kilometres. Another taxi company
has a charging structure which gives the equation C = 1.5k .
In this example there are two equations containing two variables (cost, kilometres).
Graphing these together gives
C
C=1.5k
20
C=1k+5
15
10
5
5
10
15
20
k
This is called a system of linear equations. The point where the lines intersect (cross) is the
solution of the system. From the graph it is possible to see that the fare will be the same ($15) for a
10km journey.
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There are three methods of solving a system of linear equations presented in this module;
i.
Graphically
ii.
Algebraically
iii.
Determinants
There is another method using Matrices (See Matrices module). This module will be available in
early 2014.
This module uses only systems of two equations with two unknowns.
A more complex system of linear equations would involve 3 equations with 3 variables. Each of the
methods mentioned above can possibly be used to solve more complex systems, however, the
use of determinants and matrices becomes the preferred method.
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Graphically
In this section, the two lines are drawn on the same set of axes. Unless the lines are parallel, the
lines will intersect. This point is called the point of intersection. It is a point that satisfies both
equations. One particular problem with this method is that the point of intersection may only be
located approximately.
Example:
Find the point of intersection of the two equations: y =− x + 9 and y = 2 x .
In the table below, each line is shown separately with the appropriate working. The final graph
combines the two and the point of intersection determined.
y =− x + 9
y= x − 3
x – intercept
(when y = 0)
y – intercept
(when x = 0)
x – intercept
(when y = 0)
y – intercept
(when x = 0)
x – intercept is (9,0)
y – intercept is (0,9)
x – intercept is (3,0)
y – intercept is (0,-3)
y =−0 + 9
y=9
0 =− x + 9
x=9
y
y
10
10
8
8
6
6
4
4
2
2
-2
-4
y= 0 − 3
y = −3
0= x − 3
x=3
2
4
6
8
10
x
-2
2
4
6
8
10
x
-4
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Combining the two gives:
y
10
8
6
(6,3)
4
2
-2
2
4
6
8
x
10
-4
The point of intersection is (6,3). The solution to the system of linear equations is (6,3). The point
must also satisfy both equations. To check this, the solution is substituted into both equations as
shown below.
In the equation y =− x + 9 , 6 =−3 + 9 ; and in y =
x − 3, 3 =
6 − 3 , both equations are satisfied.
Example:
Solve the system of equations; 2=
x + 5 y 10 and =
3 x − y 6 . Note these equations are presented in
general form. In the previous question the equations were presented in point-slope form.
3x − y =
6
2x + 5 y =
10
x – intercept
(when y = 0)
y – intercept
(when x = 0)
x – intercept
(when y = 0)
y – intercept is (0,2)
x – intercept is (2,0)
2× 0 + 5y =
10
y=2
2x + 5× 0 =
10
x=5
x – intercept is (5,0)
3× 0 − y =
6
3x − y =
6
x=2
y
y = −6
y – intercept is (0,-6)
y
10
10
5
5
-2
y – intercept
(when x = 0)
2
4
6
8
10
x
-2
-5
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2
4
6
8
10
x
-5
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Combining the two gives:
y
3x – y = 6
10
5
-2
2
4
6
8
-5
10
x
2x + 5y = 10
The point of intersection is approximately (2.3, 1.1). Remember, this solution is approximate. A
more accurate way of determining the point of intersection is by using algebra.
Checking the solution in the first equation: 2 x + =
5 y 10 , 2 × 2.3 + 5 × 1=
.1 10.1 , this close the
expected value of 10.
In the second equation: 3 x − y= 6 , 3 × 2.3 − 1.1
= 5.8 , this is close to the expected value of 6.
It is worth remembering that the accuracy of the point of intersection is limited by the ability to
determine the point on the graph.
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There are two systems that do not result in a single point of intersection.
The first is called the Inconsistent system. The inconsistent system results in no solution(s).
For example:
Consider the equations:
The graphs of these equations are:
=
y 3x + 1
y
10
6 x − 2 y + 11 =
0
y = 3x + 1
5
-4
-2
2
4
x
6x – 2y + 11 = 0
-5
From the graph it is clear that the lines will never intersect.
This can be shown by writing both equations in the point – slope form.
=
y 3 x + 1 is already in the point – slope form. The slope of the line is 3.
6 x − 2 y + 11 =
0 rearranged into the point – slope form becomes =
y 3x + 5.5
The slope of this line is also 3.
As both slopes are 3, the lines are parallel; there will never be a point of intersection.
The second system is the Dependent system. Like the example above, it is a result of parallel
lines. This time the lines are identical and have every point in common.
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The example below shows this.
Consider the equations:
The graph of these equations is:
y
y=
−2 x + 8
10
8 x − 4 y + 32 =
0
5
-10
-5
5
10
x
-5
-10
Writing the second equation in the point – slope form reveals what is happening here. Both
equations are y =
−2 x + 8 . This indicates that the coordinates of both lines are the same. This
explains the absence of a single solution.
Video ‘Systems of linear equations (Graph)’
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Algebraically
In this section, the solution to the system of linear equations is found algebraically. The accuracy of
the solution is improved. Within this section, there are two basic strategies that can be used. The
first is substituting one equation into the other. The second is using either addition or
subtraction to eliminate one of the variables. An earlier example is solved below.
Substitution
Example:
The two equations being considered are: y =− x + 9 and y = 2 x . With both equations are
presented in the point – slope form, the second equation can be substituted into the first.
Substituting y = 2 x into y =− x + 9 becomes 2 x =− x + 9
2 x =− x + 9
3x = 9
x=3
Now it is known that the value of x is 3, the value of y must be determined. This process is
performed by substituting x=3 into (say) the first equation. Now the value of x and y is known,
checking should be performed in the second (other) equation.
Substituting x = 3 into the first equation gives:
y =− x + 9
y =−3 + 9
y=6
The solution is ( 3, 6 )
Checking in the second equation:
y = 2x
y= 2 × 3
y=6
Correct
Solving this question was very straight-forward. The graphical method gives a good ‘picture’ of
what is happening; however, the algebraic method is often quicker.
Example:
Solve the system of equations; 2=
x + 5 y 10 and =
3x − y 6 .
This time both equations are presented in general form. To substitute one equation into the other,
one equation will require rearranging. The second equation contains a single y term, so it is the
easiest to rearrange.
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3 x − y = 6 becomes y = 3 x − 6
Substituting this into the first equation gives
2x + 5 y =
10
2 x + 5 ( 3x − 6 ) =
10
17 x − 30 =
10
17 x = 40
x
=
40
≈ 2.35 ( to 2d .p.)
17
Taking the value of x and substituting into the first equation to obtain y should be performed using
the exact value of x.
2x + 5 y =
10
40
10
+ 5y =
17
80
5=
y 10 −
17
90
5y =
17
90
y
=
≈ 1.06
85
2×
Checking in the second equation using exact values:
3x − y
40 90
−
17 85
120 90
=
−
17 85
600 − 90
=
85
510
= = 6
85
Correct
=3 ×
The check confirms the solution obtained. The solution is
1.06).
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 40 90 
 , 
 17 85 
which approximates to (2.35,
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Addition or subtraction (the elimination method)
Example:
The two equations being considered are: y =− x + 9 and y = 2 x .
The first step in this method is to write both equations in the same form, one above the other lining
up like variables. In this example this is already done.
y =− x + 9
y = 2x
The aim of this step is to eliminate one of the variables. If the equations were subtracted, the y
variable would be eliminated.
y =− x + 9
− y=
2x
0 =−3 x + 9 → x =3
The next steps are the same as previously performed.
Substituting x = 3 into the first equation gives:
y =− x + 9
y =−3 + 9
y=6
The solution is ( 3, 6 )
Checking in the second equation:
y = 2x
6= 2 × 3
Correct
Example:
The two equations being considered are: x + 2 y =
7 and y − x =
8.
The first step in this method is to write both equations in the same form.
x + 2y = 7 → x + 2y = 7
y −=
x 8
→ − x +=
y 8
The x term only differs by sign, so if the two equations were added (x + -x = 0).
+
x + 2y =
7
−x + y =
8
3 y = 15 → y = 5
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Substituting y = 5 into the first equation gives:
x + 2y =
7
x + 2×5 =
7
x = −3
Checking in the second equation:
y−x
= 5 − ( −3)
=8
Correct
Example:
Find the point of intersection of the lines:
y =−2 x − 7 and 2 y − x =
1
Write the equations in the same form.
y=
−2 x − 7 → y + 2 x =
−7
2 y − x= 1 → 2 y − x= 1
In this example, addition or subtraction cannot take place until the x or y coefficients are the same
(except for sign). To achieve this, the first equation is multiplied by 2. After this the equations can
be subtracted to eliminate the y variable.
-7 )
( y + 2x =
=
2y - x 1
× 2 →
-14
2 y + 4x =
→ - =
2y - x 1
-15 → x =
-3
5x =
Substituting x = -3 into the first equation gives:
y=
−2 x − 7
y =−2 × ( −3) − 7
y= 6 − 7
y = −1
Checking in the second equation:
2y − x
= 2 × ( −1) − ( −3)
=1
Correct
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Example:
Find the point of intersection of the lines:
3x + 5 y =
−9 and 4 x − 3 y =
17
The equations are in the same form.
This question requires the multiplication of both equations in order to eliminate a variable. To
eliminate x, the first equation is multiplied by 4 and the second by 3. To eliminate y, the first
equation is multiplied by 3 and the second by 5.
Eliminating x by subtracting:
−9 )
( 3x + 5 y =
y 17 )
( 4 x − 3=
12 x + 20 y =
−36
× 4 →
y 51
× 3 → − 12 x − 9=
29 y =
−87 → y =
−3
Substituting y = -3 into the first equation gives:
3x + 5 y =
−9
3 x + 5 ( −3) =
−9
3x = 6
x=2
Checking in the second equation:
4x − 3y
=4 × ( 2 ) − 3 ( −3)
= 8+9
= 17
Correct
Example:
Solve the system of equations; 2=
3x − y 6 .
x + 5 y 10 and =
The equations are in the same form. If the second equation is multiplied by 5, then the y variable
can be eliminated by addition.
2=
x + 5 y 10
y
( 3x −=
6)
→
×5
→ +
2=
x + 5 y 10
15 x − 5=
y 30
17 x
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= 40
40
x=
17
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Substituting
x=
40
17
into the first equation gives:
Checking in the second equation:
2x + 5 y =
10
40
2× + 5y =
10
17
90
5y =
17
90 18
=
y =
85 17
3x − y
40 90
−
17 85
600 − 90
=
85
510
= = 6
85
Correct
=×
3
Video ‘Systems of linear equations (Algebraic)’
Video ‘Systems of linear equations (Application)’
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Using determinants
In this section, a totally different method of solving Systems of Linear Equations will be covered.
The theory behind the method used will not be covered, but the process of how determinants can
be used will be presented as a sequence of steps to follow.
It is necessary to be able to solve a determinant of the second order. An example of a second
order determinant is:
det
a b
= ad − bc
c d
Evaluating a second order determinant is performed as below.
det
4 −1
3
7
= 4 × 7 − ( −1) × 3 = 28 + 3 = 31
Example:
Solve this system of linear equations using determinants: y =
−3 x + 10 and y = 2 x .
The first step is to write both equations in the form of
ax + by =
c
y=
−3 x + 10 → 3x + 1 y =
10
y 2x
1y 0
=
→ − 2 x +=
↑
↑
x
y
coefficients
↑
constants
From this a 2 x 2 (square) matrix (A) will be formed using only the coefficients of the variables x
and y.
 3 1
A=

 −2 1
This matrix is modified to produce two new matrices. The matrix A* is formed by replacing the x
coefficients with the constants. The matrix A** is formed by replacing the y coefficients with the
constants. Altogether there are three matrices.
 3 1
10 1
 3 10 
=
=
A =
A*
A*
*

 0 1
 −2 0 
 −2 1




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The solution of the system of linear equations is given by:
10 1
=
x
y
=
det A*
=
det A
0
3
−2
1
10 × 1 − 1 × 0 10
=
= = 2
1 3 × 1 − 1 × ( −2 ) 5
1
3
10
det A* * −2
=
3
det A
−2
0 3 × 0 − 10 × ( −2 ) 20
=
= = 4
1
3 × 1 − 1 × ( −2 )
5
1
The solution (point of intersection) is (2,4). This can be checked by substituting into the original
equations as in previous methods.
This method of solving linear equations is called Cramer’s Rule. Once the method is practised,
solving linear equations is fairly straightforward.
Example:
Find the point of intersection of the lines:
3x + 5 y =
−9 and 4 x − 3 y =
17
The first step is to write both equations in the form of
ax + by =
c
3x + 5 y =
−9
4x − 3y =
17
From this a square matrix (A) will be formed using only the coefficients of the variables x and y.
3 5 
A=

 4 −3
Modify this matrix to produce two new matrices. The matrix A* is formed by replacing the x
coefficients with the constants. The matrix A** is formed by replacing the y coefficients with the
constants. Altogether there are three matrices.
3 5 
 −9 5 
 3 −9 
=
A =
A*
=
A*
*

17 −3
 4 17 
 4 −3




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The solution of the system of linear equations is given by:
−9 5
det A* 17 −3
x
=
=
=
3 5
det A
4 −3
17
( −9 ) × ( −3) − 5 ×=
3 × ( −3) − 5 × 4
−58
= 2
−29
3 −9
det A** 4 17 3 × 17 − ( −9 ) × 4 87
y=
=
=
=
= −3
3 5
det A
3 × ( −3) − 5 × 4 −29
4 −3
The solution (point of intersection) is (2,-3). This can be checked by substituting into the original
equations as in previous methods.
Example:
Solve the system of equations; 2=
x + 5 y 10 and =
3x − y 6 .
Both equations are in the correct form.
From this the matrix A =
2x + 5 y =
10
3x − y =
6
2
5
3 −1
From this the matrices A* and A** are formed.
=
A
The solution is:
2 5
10 5
2 10
=
A*
=
A**
3 −1
6 −1
3 6
10 5
6 −1 10 × ( −1) − 5 × 6 −40 40
det A*
=
x
=
=
= =
2 5
2 × ( −1) − 5 × 3 −17 17
det A
3 −1
=
y
det A**
=
det A
2 10
3 6
−18 18
2 × 6 − 10 × 3
=
= =
2 5
2 × ( −1) − 5 × 3 −17 17
3 −1
 40 18 
, 
 17 17 
The solution (point of intersection on the graph) is 
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Video ‘Systems of linear equations (Determinants)’
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Activity
1.
Evaluate the determinants
(a)
(b)
4 6
10 −5
0 5
3 5
(c)
−2.1 −1.5
8
2.
(a)
(d)
y= x + 4
y = 3x
(b)
4x + y =
−19
(d)
4x + y =
−19
(f)
8
−2 x + y =
7 x − 5 y + 39 =
0
14 x + 3 y + 13 =
0
(h)
4x + y =
−19
−2 x + y =
8
(j)
5.2 x + 3.4 y =
15.8
2.6 x − 4.2 y =
25.6
5.2 x + 3.4 y =
15.8
2.6 x − 4.2 y =
25.6
5 x + 11 y =
32
0.5 x + y − 4 =
0
Algebraic method
(l)
4x + 3y =
6
=
y 2x − 2
Determinant method
(m)
−3 x + 2 y =
7
x − 3y =
1
Algebraic method
Algebraic method
(k)
−3 x + 2 y =
7
x − 3y =
1
Algebraic method
Algebraic method
4x + 3y =
6
=
y 2x − 2
2x + y =
10
x− y =
5
Graphical method
Algebraic method
(i)
x +1
Graphical method
Graphical method
(g)
x
Solve the system of linear equations by the method indicated.
8
−2 x + y =
(e)
x
5
Graphical method
(c)
x −1
Determinant method
(n)
Determinant method
5 x + 11 y =
32
0.5 x + y − 4 =
0
Determinant method
(o)
A customer purchases four chocolate bars and one ice cream and pays $8.70; another
customer purchases one chocolate bar and two ice creams and pays $6.55, what is
the unit price of chocolate bars and ice creams?
(p)
The sum of two numbers is 4, where the difference of the two numbers is 9. Determine
the two numbers.
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(q)
Two cables support a weight of 120N as shown in the diagram below.
T2
T1
120N
10N
The equations below apply to the direction indicated;
120
Vertically - 0.71T1 + 0.866T2 =
10 + 0.5T2
Horizontally - 0.71T=
1
Calculate the tensions T1 and T2
(r)
A volume of 6% solution is mixed with a different volume of 15% solution to obtain
200mL of a 10% solution. Calculate the volumes of each of the original solutions.
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Answers
1.
Evaluate the determinants
(a)
(c)
2.
4 6
= 4×5 − 6×3 = 2
3 5
(b)
−2.1 −1.5
= ( −2.1) × 5 − ( −1.5 ) × 8 = 1.5
8
5
(d)
10 −5
0
5
= 10 × 5 − ( −5 ) × 0 = 50
x −1
x
=
x2 − 1 − x2 =
−1
( x − 1)( x + 1) − x 2 =
x
x +1
Solve the system of linear equations by the method indicated.
y= x + 4
y = 3x
(a)
2x + y =
10
x− y =
5
(b)
Graphical method
Graphical method
y
y
10
10
(2,6)
5
5
(5,0)
-10
-5
5
10
x
-5
5
-5
-5
-10
-10
4x + y =
−19
−2 x + y =
8
(c)
-10
Graphical method
y
y
-5
10
10
5
5
(-4.5,-1)
x
−3 x + 2 y =
7
x − 3y =
1
(d)
Graphical method
-10
10
5
10
x
-10
-5
(-3.3,-1.5)
-5
-5
-10
-10
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5
10
x
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4x + y =
−19
(e)
7
−3 x + 2 y =
1
x − 3y =
(f)
8
−2 x + y =
Algebraic method
Algebraic method
-19
4x + y =
--2 x + y =
8
−3 x + 2 y = 7 →
x − 3 y = 1 → ×3 →
= -27
x = -4.5
4 × -4.5 + y = -19
-18 + y =
-19
y = -1 Solution is (-4.5,-1)
Check in equation 2
-2 x + y
=-2 × ( -4.5 ) + ( -1)
=8
Correct
7 x − 5 y + 39 =
0
14 x + 3 y + 13 =
0
Algebraic method
7 x − 5 y + 39 =0 → ×2 → 14 x − 10 y + 78 =0
14 x + 3 y + 13 =0 → − 14 x + 3 y + 13 =0
− 13 y + 65 =
0
y=5
7 x − 5 y + 39 =0 → 7 x − 5 × 5 + 39 =0
7 x = −14
x = −2
Solution is ( −2 ,5 )
Checking
14 x + 3 y + 13
= 14 × ( −2 ) + 3 × 5 + 13
=0
− 3x + 2 y = 7
3x − 9 y = 3
10
− 7y =
10
y= −
7
 10 
−3 x + 2 y = 7 → − 3 x + 2 ×  −  = 7
 7
20
− 3 x= 7 +
7
23
x= −
7
 23 10 
Solution is  − , − 
7
 7
Checking
x − 3y
23
−10
=− − 3 ×
7
7
−23 + 30
= = 1
7
Correct
6x
(g)
+
5.2 x + 3.4 y =
15.8
2.6 x − 4.2 y =
25.6
(h)
Algebraic method
5.2 x + 3.4 y 15.8
5.2 x + 3.4 y 15.8
=
=
2.6 x − 4.=
2 y 25.6 × 2 → − 5.2 x − 8.=
4 y 51.2
11.8 y = −35.4
y = −3
5.2 x + 3.4=
y 15.8 → 5.2 x + 3.4 × ( −3=
) 15.8
5.2 x = 26
x=5
Solution is ( 5,−3)
Checking
2.6 x − 4.2 y
= 2.6 × 5 − 4.2 × ( −3)
= 25.6
Correct
Correct
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(i)
4x + 3y =
6
5 x + 11 y =
32
0.5 x + y − 4 =
0
(j)
=
y 2x − 2
Algebraic method
Algebraic method
5 x + 11 y = 32 →
4x + 3y =
6
y 2x − 2
=
×10
0.5 x + y − 4 = 0 → − 5 x + 10 y = 40
by substitution
4 x + 3( 2 x − 2) =
6
y = −8
10 x = 12
6
x=
5
(k)
6
4 x + 3 y =6 → 4 ×   + 3 y =6
5
6
3y =
5
2
y=
5
6 2
Solution is  , 
5 5
Checking
=
y 2 x − 2?
2
6
=2× − 2
5
5
2 2
=
5 5
Correct
4x + y =
−19
5 x + 11 y= 32 → 5 x + 11× ( −8 )= 32
5 x = 120
x = 24
Solution is ( 24 , −8 )
Checking
0.5 x + y − 4
= 0.5 × 24 + ( −8 ) − 4
=0
Correct
4x + 3y =
6
=
y 2x − 2
(l)
8
−2 x + y =
Determinant method
3 5 
 −9 5 
 3 −9 
=
A =
A*  =
A** 



 4 −3
17 −3
 4 17 
det A* 27 − 85 −58
=
=
= = 2
x
det A −9 − 20 −29
det A** 51 + 36 87
=
=
= −3
y=
−29
−29
det A
The solution is ( 2 , −3)
(m)
5.2 x + 3.4 y =
15.8
2.6 x − 4.2 y =
25.6
5 x + 11 y = 32
Determinant method
y 6
4 x + 3=
y = 2x − 2
→ 4 x + 3=
y 6
→ 2x − y = 2
4 3 
6 3 
4 6
=
A =
A*  =
A** 



 2 −1
 2 −1
2 2
det A* −6 − 6 −12 6
=
=
= =
x
det A −4 − 6 −10 5
det A** 8 − 12 −4 2
=
=
= =
y
−10
−10 5
det A
6 2
The solution is  , 
5 5
(n)
Determinant method
5.2 3.4 
15.8 3.4 
5.2 15.8 
=
A =
 A*  25.=
 A**  2.6 25.6 
−
−
2
.
6
4
.
2
6
4
.
2






det A* −66.36 − 87.04 −153.4
=
= = 5
x =
−21.84 − 8.84 −30.68
det A
det A** 133.12 − 41.08 92.04
=
=
= −3
y=
−30.68
−30.68
det A
The solution is ( 5, −3)
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5 x + 11 y =
32
0.5 x + y − 4 =
0
Determinant method
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5 x + 11 y = 32
→ 5 x + 11 y = 32
0.5 x + y − 4 = 0 → 0.5 x + y = 4
 5 11
32 11
 5 32 
A =
=
 A*  4=
 A** 0.5 4 
0
5
1
1
.






det A* 32 − 44 −12
x
=
=
= = 24
5 − 5.5 −0.5
det A
4
det A** 20 − 16
y=
=
=
= −8
det A
−0.5
−0.5
The solution is ( 24 , −8 )
(o)
A customer purchases four chocolate bars and one ice cream and pays $8.70; another
customer purchases one chocolate bar and two ice creams and pays $6.55, what is
the unit price of chocolate bars and ice creams?
Let the price of the chocolate bar be C, let the price of the ice-cream be I.
C + I 8.7 and C
=
+ 2 I 6.55
The two equations formed are: 4=
Solving by algebra:
4C +=
I 8.7
→
4C +=
I 8.7
×4
(p)
C +=
2 I 6.55 → − 4C +=
8 I 26.2
− 7I =
−17.5
I = 2.5
4C + 2.5 =
8.7
4C = 6.2
C = 1.55
Solution is
Chocolate Bars cost $1.55 and Ice Creams $2.50
Check
C + 2I
= 1.55 + 2 × 2.5
= 6.55
Correct
The sum of two numbers is 4, where the difference of the two numbers is 9. Determine
the two numbers.
Let the first number be A, and the second B, the two equations formed are:
A=
+ B 4 and A=
−B 9
Solving by algebra:
A+ B =
4
− A− B =
9
2 B = −5
B = −2.5
A + B = 4 → A + ( −2.5 ) = 4
A = 6.5
Solution is A = 6.5, B = −2.5
Check
A− B
= 6.5 − ( −2.5 )
=9
Correct
(q)
Two cables support a weight of 120N as shown in the diagram below.
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T2
T1
120N
10N
The equations below apply to the direction indicated;
120
Vertically: 0.71T1 + 0.866T2 =
10 + 0.5T2
Horizontally: 0.71T=
1
Calculate the tensions T1 and T2
Solving by Determinants
0.71T1 + 0.866T2 =
120
0.71T1 − 0.5T2 =
10
0.71 0.866 
120 0.866 
0.71 120 
=
A =
A*  =
A** 



0.71 −0.5 
 10 −0.5 
0.71 10 
−60 − 8.66
−68.66
det A*
=
= = 70.8
T1 =
det A −0.355 − 0.615 −0.97
det A** 7.1 − 85.2 −78.1
=
=
= = 80.51
T2
−0.97
−0.97
det A
The solution
=
is T1 70
=
.8 N T2 80.51N
(r)
A volume of 6% solution of weedkiller is mixed with a different volume of 15% solution
to obtain 200mL of a 10% solution. Calculate the volumes of each of the original
solutions.
Let the volume of the 6% solution be x, and the volume of 15% solution be y.
200
This gives an equation: x + y =
Consider the amount of weedkiller in each solution:
0.06 x + 0.15 y =
0.1× 200
0.06 x + 0.15 y =
20
Solving by Algebra:
×0.06
x=
+ y 200 →
0.06 x + 0.06
=
y 12
0.06 x + 0.15 y =
20 → − 0.06 x + 0.15 y =
20
− 0.09 y =
−8
y = 88.9
x + y= 200 → x + 88.9= 200
x = 111.1
Solution is (111.1,88.9 )
Check
0.06 x + 0.15 y
= 0.06 × 111.1 + 0.15 × 88.9
= 20
Correct
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