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Olympic College Topic 15 Solving System of Linear Equations

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Topic 15 –Solving System of Linear Equations

Introduction: A “Linear Equation” is an equation that contains two variables, typically x and

y that when they are graphed on a cartesian grid make a straight line.

There are a number of different forms of linear equation but the two most common

formats are the”Slope Intercept Form” and the “Standard Form”.

The Slope Intercept form is written as y = mx + b where m is the slope of the line and b is the

y-intercept. For example y = 4x – 1 is a linear equation written in its slope intercept form.

The Standard Form for a linear equation is ax + by = c where a is called the coefficient of x and b is

called the coefficient of y and c is called the constant term.

For example, 2x – 5y = 10 is a linear equation written in its standard form.

1. Solving a System of Equations Graphically.

Definition: A system of linear equations is when you are given two linear equations.

Solving a system of linear equations is the process of trying to find a solution that satisfies

both equations. When you try to solve a system of equations there are three distinct

possibilities.

The diagrams below show visually the three possibilities.

Situation 1: Situation 2: Situation 3:

In the above situation the In the above situation the two In the above situation the

two lines cross at one point. lines are parallel and so never two lines are identical and

In this situation we say that cross. In this situation we say In this situation we say that

the system is Consistent that the system is Inconsistent the system is Dependant

and has one solution. and has no solutions. Has a infinite solutions.

x

y

x

y

x

y


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It is possible to find the number of solutions to a system of equations by expressing each linear equation

in its slope intercept form y = mx + b and then comparing the two equations.

If both lines have different slopes the system of equations has a one solution and is called

“Consistent”.

If both lines have the same slope but different y-intercepts then the system of equations has a no

solutions and is called “Inconsistent”.

If both lines are identical then there are an infinite number of solutions and the system of

equations is called “Dependant”.

Example 1: For the following system of equations state the type of system of equation we have and the

number of solutions it will have.

(a) x + 2y = 6 and 2x + y = 4

(b) ¼x – y = 4 and ½ x – 2y = – 6

(c) 6x + 3y = 12 and y = – 2x + 4

Solution (a) x + 2y = 6 2x + y = 4

2y = – x + 6 y = – 2x + 4

y = – ½ x + 3

These equations have different slopes so we have a Consistent system of equations with

one solution,.

Solution (b) ¼x – y = 4 ½ x – 2y = – 6

– y = ¼x + 4 – 2y = – ½ x – 6

y = ¼x – 4 y = ¼ x + 3

These equations have the same slope but different y-intercepts so we have an Inconsistent

system of equations no solutions.

Solution (c) 6x + 3y = 12 y = – 2x + 4

3y = – 6x + 12

y = – 2x + 4

These equations are identical so we have a Dependant system of equations with an infinite

number of solutions.


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In order to solve a system of equations graphically you draw both lines on a Cartesian grid and then read

off the point of intersection if it exists.

Example 1: Solve the system of equations y = 2x + 4

y = 3x – 6

Solution: For the line y = 2x + 4

To find the y-intercept, make x = 0 and solve

y = 2(0) + 4

y = 4 So the line cuts the y-axis at 4.

To find the x-intercept, make y = 0 and solve

0 = 2x + 4

– 4 = 2x

= x So the line cuts the x-axis at – 2.

For the line y = – 3x – 6

To find the y-intercept, make x = 0 and solve

y = – 3(0) – 6

y = – 6 So the line cuts the y-axis at – 6.

To find the x-intercept, make y = 0 and solve

y = – 3x – 6

0 = – 3x – 6

6 = – 3x

– 2 = x So the line cuts the x-axis at – 2.

The two lines intersect at the unique point (– 2 , 0) in this situation the system of equations is

said to be “Consistent” and the solution to the system of equations is the point(– 2 , 0) or

it can also be expressed as x = – 2 and y = 0.

x

y

0

y = 2x+4

y = – 3x – 6


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Example 2: Solve the syatem of equations y = 2x + 4

2y – 4x = 4

For the line y = 2x + 4

To find the y-intercept, make x = 0 and solve y = 2(0) + 4 = 4

So the line cuts the y-axis at 4.

To find the x-intercept, make y = 0 and solve 0 = 2x + 4

– 4 = 2x

– 2 = x

So the line cuts the x-axis at – 2.

For the line 2y – 4x = 4

To find the x-intercept, make y = 0 and solve 2(0) – 4x = 4

– 4x = 4

x = – 1

So the line cuts the x-axis at – 1

To find the y-intercept, make x = 0 and solve 2y – 4(0) = 4

2y = 4

y = 2

So the line cuts the y-axis at 2

The two lines are parallel and so will never meet, this is interpreted as there is no solution

to the system of linear equations and in these situations we say that the system of equations

is “Inconsistent”.

x

y

0

y = 2x+4

2y – 4x = 4


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Example 3: Solve the syatem of equations y = 2x + 4

2y – 4x = 8

For the line y = 2x + 4

To find the y-intercept, make x = 0 and solve y = 2(0) + 4 = 4

So the line cuts the y-axis at 4.

To find the x-intercept, make y = 0 and solve 0 = 2x + 4

– 4 = 2x

– 2 = x

So the line cuts the x-axis at – 2.

For the line 2y – 4x = 8

To find the x-intercept, make y = 0 and solve 2(0) – 4x = 8

– 4x = 8

x = – 2

So the line cuts the x-axis at – 2

To find the y-intercept, make x = 0 and solve 2y – 4(0) = 8

2y = 8

y = 4

So the line cuts the y-axis at 4

The two lines are identical and so every point on the line is a potential solution, this is

interpreted as there are an infinite number of solutions to the system of linear equations

and in these situations we say that the system of equations is “Dependant”.

x

y

0

y = 2x+4 2y – 4x = 8


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Exercise 1

1. For the following system of equations state the type of system of equation we have and the number

of solutions it will have.

(a) 3x + 5y = 15 and 4x + 3y = 12

(b) ¼x – ½y = 2 and x – y = – 2

(c) 6x + 2y = 12 and y = – 3x + 6

(d) 2x – ½y = 5 and 4x – y = 1

(e) x + 2y = 6 and 4y = 2x + 12

2. Solve the following system of equations graphically.

(a) 3x + 5y = 13 and 4x + 3y = 10

(b) ¼x – ½y = 2 and x – y = 6

(c) 2x + 7y = 14 and 4x +14y = 10

(d) 4x – 3y = 24 and 8x – 6y = 48

(e) 2x + y = 0 and 3y = 2x – 14


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2. Solving a System of Equations Algebraically

There are two common algebraic methods used to solve a system of equations. They are:-

A. The Substitution method

B. The Addition Method

A. Solving A System of Equations Using The Substitution Method

This method uses the technique of rearranging one of the two linear equations into the form y = … or

x = ….. and then substituting this new equation into the second equation. The result of doing this is that

we will eliminate one of the two variables x or y and will be left with a linear equation in only one

variable which we then solve in the usual way.

Example 1: Solve the system of equations y = 2x – 13

– 4x – 7 = 9y

Solution. Equation 1: y = 2x – 13

Equation 2: – 4x – 7 = 9y

We will use equation 1 and substitute the y = 2x – 13 into equation 2 to get the following:

– 4x – 7 = 9y

– 4x – 7 = 9(2x – 13)

– 4x – 7 = 18x – 117

– 4x = 18x – 110

– 22x = – 110

x = 5

We now substitute the x = 5 into equation 1:

y = 2x – 13

y = 2(5) – 13

y = – 3.

The solution to the system of equations is the point (5, – 3).


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Example 2: Solve the system of equations 2x + 3y = 7

6x – y = 1

Solution. Equation 1: 2x + 3y = 7

Equation 2: 6x – y = 1

We will use equation 1 and rearrange it into the form y = …….

Re-arrange 6x – y = 1

– y = – 6x + 1

y = 6x – 1

We now substitute the equation y = 6x – 1 into equation 1 to get the following:

2x + 3y = 7

2x + 3(6x – 1) = 7

2x + 18x – 3 = 7

20x – 3 = 7

20x = 10

x = 0.5

We now substitute the x = 0.5 into equation 2: y = 6x – 1

y = 6(0.5) – 1

y = 2

The solution to the system of equations is the point(0.5, 2).

Example 3. Solve the system of equations y = – 2x + 5

6x + 3y = 0

Solution. Equation 1: y = – 2x + 5

Equation 2: 6x + 3y = 0

We now substitute the equation y = – 2x + 5 into equation 2 to get the following:

x + 3y = 0

6x + 3(– 2x + 5 ) = 0

6x – 6x + 15 = 0

15 = 0

This is clearly impossible, we interpret any situation like this where all the variables

disappear and we are left with an impossible statement as an Inconsistent system and in

these there will be no solutions.


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Example 4. Solve the system of equations 3y + x = 10

3x + 4y = 15

Solution. Equation 1: 3y + x = 10


We will use equation 1 and rearrange it into the form x = ……

3y + x = 10

x = – 3y + 10

We now substitute the equation x = – 3y + 10 into equation 2 to get the following:

3x + 4y = 15

3(– 3y + 10) + 4y = 15

– 9y + 30 + 4y = 15

– 5y + 30 = 15

– 5y = – 15

y = 3

We now substitute the y = 3 into the equation x = – 3y + 10

x = – 3(3) + 10

x = 1

The solution to the system of equations is the point(1, 3).

Example 5: Solve the system of equations 2x + y = 3

3y = 9 – 6x

Solution : Equation 1: 2x + y = 3

Equation 2: 3y = 9 – 6x

Choose equation 2 and rearrange in the form y = …..

3y = 9 – 6x

y = 3 – 2x

We now substitute the equation y = 3 – 2x into equation 1: to get the following:

2x + y = 3

2x + 3 – 2x = 3

3 = 3

When we get a situation like this where the variables all disappear and we are left with a

true statement we interpret such situations as a Dependant system of equations and in these

situations we will have solutions.


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Example 6 Solve the system of equations x = y + 4

3x + 7y = – 18

Solution : Equation 1: 4x – 3y = 6


Re-arrange 2x + 4y = 5

2x = – 4y + 5

x = – 2y + 2

5

Use equation 2 and substitute the y = 6x – 1 into equation 1 to get the following:

2x + 3y = 7

2x + 3(6x – 1) = 7

2x + 18x – 3 = 7

20x – 3 = 7

20x = 10

x = 0.5

We can the substitute the x = 0.5 into equation 2: y = 6x – 1 = 6(0.5) – 1 = 2

The solution to the system of equations is the point(0.5, 2).

Example 7 Solve the system of equations 0.1x + 0.02y = 0.9

0.2x + 0.01y = 1.2

Solution: First we change the equations so that all the coefficients are whole numbers.

In those situations where the coefficients are decimals you can do this by multiplying all

the terms by 10, 100 or 1000 depending on the number of largest number of decimal

places the coefficients have. In this case it is two decimal places so we will multiply the

equations by 100.

Equation 1: 0.1x + 0.02y = 0.9 x 100 10x + 2y = 90

Equation 2: 0.2x + 0.01y = 1.2 x 100 20x + y = 120

Rearrange 20x + y = 120 to get y = – 20x + 120

We now substitute y = – 20x + 120 into the equation 10x + 2y = 90

10x + 2(– 20x + 120) = 90

10x – 40x + 240 = 90

– 30x + 240 = 90

– 30x = – 150

x = 5

We can the substitute the x = 5 into the equation y = – 20x + 120 = – 20(5) + 120 = 20

The solution to the system of equations is the point(5, 20).


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Example 8 Solve the system of equations

x +

y =

x +

y =


In those situations where the coefficients are fractions you can do this by multiplying all

the terms in each equation by the lowest common multiple (LCM) of all the denominators.

In this case, for equation 1 it is the LCM of 3,4 and 8 which is 24 and for equation 2 it is

the LCM of 2 and 5 which is 10.

Equation 1:

x +

y =

x 24 8x + 18y = – 39

Equation 2:

x +

y =

x 10 5x + 4y = – 28

Rearrange 5x + 4y = – 28 to get

4y = – 5x – 28

y =

We now substitute y =

into the equation 8x + 18y = – 39

8x + 18(

) = – 39

8x –

x – 126 = – 39

–

x – 126 = – 39

–

x = 87

x = – 6

We can the substitute the x = – 6 into the equation y =

y =

y =

y =

The solution to the system of equations is the point (– 6,

).


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Exercise 2A

Solve the following system of equations using the Substitution Method.

1. 2x + 3y = 11 2. 4x – 4y = 0

4x – 2y = – 2 y = 1

3. x + 5y = 0 4. 2x – 5y = 10

3x – 4y = 38 6x – 15y = 30

5. x + 3y = – 2 6. x – 4y = 10

5x + 15y = 0 3x – 2y = 10

7. 4a + 7b = 54 8. 2x – 3y = 1

2a – 3b = 14 8x – 12y = 4

9. 3x + 4y = 12 10. 2a – 5b = 10

6x + 8y = 24 3a – b = 2

11. 0.3x + 0.4y = 1.1 12.

=

0.02x + 0.01y = 0.04

=


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B. Solving A System of Equations Using The Addition Method

The Addition Method requires us to rearrange the two equations so that they both contain the same

number of either x or y’s but one must also be negative. Once we have achieved this goal we can then

add the two equations together and so eliminate this variable. We will then be left with a linear

equation in only one variable that we can then solve in the usual way.

Example 1: Solve the system of equations x – 2y = 6

x + 2y = 8

Solution: Equation 1: x – 2y = 6

Equation 2: x + 2y = 8

These two equations are already in the correct form as they have a +2y and the other – 2y

We now add he two equations together and eliminate the y variable.

x – 2y = 6

x + 2y = 8

2x = 14

x = 7

We now know that x = 7 to find the corresponding value of y we substitute x = 7 into the

equation x + 2y = 8 to get

x + 2y = 8

7 + 2y = 8

2y = 1

y = ½

So the solution is (7, ½)

Note: You can always check that your solution is correct by substituting it into both equations and

checking that they work.

Using x = 7 and y = ½ in Equation 1 we get x – 2y = 6

7 – 2(½) = 6

7 – 1 = 6

6 = 6 (checks)

Using x = 7 and y = ½ in Equation 2 we get x + 2y = 8

7 + 2(½) = 8

7 + 1 = 8

8 = 8 (checks)


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Example 2: Solve the system of equations 2x – 3y = 4

2x + y = – 4

Solution: Equation 1: 2x – 3y = 4

Equation 2: 2x + y = – 4

In this example we have to change one of the equations we can either change equation 2 by

multiplying it by – 1 and by doing this we get the – x term that we need. On the other hand

we could multiply equation 2 by 3 in order to get the + 3y term that we need. In this

situation we will multiply equation 2 by 3.

2x – 3y = 4 2x – 3y = 4

2x + y = – 4 multiply by 3 6x + 3y = – 12 (add)

8x = – 8

x = – 1

Use x = – 1 in the equation 2x + y = – 4

2(– 1) + y = – 4

– 2 + y = – 4

y = – 2

So the Solution is (– 1, – 2)

Example 3: Solve the system of equations x – y = 2

3x – 3y = 0

Solution: Equation 1: x – y = 2

Equation 2: 3x – 3y = 0

In this situation we will multiply equation 1 by – 3 to get the + 3y that we need.

x – y = 2 multiply by – 3 –3x +3y = – 6

3x – 3y = 0 3x – 3y = 0 (add)

0 = – 6

In this situation we have added the equations together and all the variables have been

eliminated. The result is that we are left with the impossible result that 0 = – 6.

We interpret this situation as the system of equations is Inconsistent and so there are no

solutions.


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Example 4: Solve the system of equations x – 3y = 11

–3x = – 5y – 17


Equation 2: –3x = – 5y – 17

The first step is to get both equations into their standard form ax + by = c.

We have to rearrange equation to from –3x = – 5y – 17 into –3x + 5y = – 17


Equation 2: –3x +5y = – 17

We can now focus in trying to eliminate one of the variables by matching them up.

When you look at the two equations we cannot just change one we need to change both

equations in order to get the variables to match.

In this question if we focus on the x terms we have + 2x and – 3x we can change both into

+ 6x and – 6x by multiplying equation 1 by 3 and equation 2 by 2.

2x – 3y = 11 multiply by 5 10x – 15y = 55

–3x + 5y = – 17 multiply by 3 –9x + 15y = – 51 (add)

x = 4

We now use x = 4 in the equation 2x– 3y = 11

2(4) – 3y = 11

8 – 3y = 11

– 3y = 3

y = – 1

So the solution is (4, – 1)

Example 5: Solve the system of equations 2x – 3y = 3

– 4x + 6y = – 6


Equation 2: – 4x + 6y = –6

In this question we can change the – 3y into the – 6y that we need.

2x – 3y = 3 multiply by 2 4x – 6y = 6

4x + 5y = 6 4x + 6y = – 6 (add)

0 = 0

In this situation we have added the equations together and all the variables have been

eliminated. The result is that we are left with the true statement that 0 = 0. We interpret this

situation as we have a Dependant system of equations and that there will be solutions.


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Example 6: Solve the system of equations 5x = 2y – 4

3x – 5y = 6

Solution: Equation 1: 5x = 2y – 4


The first step is to get both equations into their standard form ax + by = c.

We have to rearrange equation to from x = 2y – 4 into x – 2y = – 4

Equation 1: 5x – 2y = – 4



When you look at the two equations we cannot just change one we need to change both

equations in order to get the variables to match.

In this question if we focus on the y terms we have – 2y and – 5y we can change both into

+ 10x and – 10x by multiplying equation 1 by – 5 and equation 2 by 2.

We now use these two equations

5x – 2y = – 4 multiply by – 5 – 25x + 10y = 20

3x – 5y = 6 multiply by 2 6x – 10y = 12 (add)

– 19x = 32

x = 19

32

We now use x = 19

32 in the equation 3x – 5y = 6

3(19

32 ) – 5y = 6

19

96 – 5y = 6

– 5y = 6 + 19

96

– 5y = 19

210

y = 19

42

So the Solution is (19

32 ,

19

42 ).


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Example 7 Solve the system of equations 0.2x + 0.12y = 0.64

0.1x – 2.4y = 7.7


In those situations where the coefficients are decimals you can do this by multiplying all

the terms by 10, 100 or 1000 depending on the number of largest number of decimal

places the coefficients have in each equation. In this case it is two decimal places for

equation 1 and 1 decimal place for equation 2.

Equation 1: 0.2x + 0.12y = 0.64 x 100 20x + 12y = 64

Equation 2: 0.1x – 2.4y = 7.7 x 10 x – 24y = 77


In this question if we focus on the y terms we have + 12y and – 24y we can change both

into + 12y and – 24y by multiplying equation 1 by 2.


20x + 12y = 64 multiply by2 40x + 24y = 128

x – 24y = 77 x – 24y = 77 (add)

41x = 205

x = 5

We now use x = 5 in the equation 20x + 12y = 64

20(5) + 12y = 64

100 + 12y = 64

12y = – 36

y = – 3

So the Solution is(5, – 3)


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Example 8 Solve the system of equations

x

y =

x +

y =


In those situations where the coefficients are fractions you can do this by multiplying all

the terms in each equation by the lowest common multiple (LCM) of all the denominators.

In this case, for equation 1 it is the LCM of 2,5 and 10 which is 10 and for equation 2 it is

the LCD of 4 and 8 which is 8.

Equation 1:

x

y =

x 10 5x – 6y = – 1

Equation 2:

x +

y =

x 8 5x + 6y = 11


In this question he equations are all ready to be added and we need to do no more.


5x – 6y = – 1

5x + 6y = 11 (add)

10x = 10

x = 1


5(1) + 6y = 11

5 + 6y = 11

6y = 6

y = 1

So the Solution is(1, 1)


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Exercise 2B

Solve the following system of equations using the Addition Method.

1. x + 3y = 10 2. x – y = 0

3x – 3y = – 6 2x – 3y = – 1

3. 2x + 5y = 30 4. 6x – 15y = 30

3x – 4y = 22 2x – 5y = 10

5. 2x + 5y = – 5 6. 3x – 4y = 14

10x + 25y = 0 2x – 2y = 8

7. 4a + 7b = 54 8. 2x – 3y = 1

2a – 3b = 14 8x – 12y = 4

9. 3x + 4y = 12 10. 2a – 3b = 6

6x + 8y = 24 3a – 2b = 4

11. 0.6x + 0.8y = 2.2 12.

=

0.01x + 0.005y = 0.02

=


Page | 20

3. Application Problems involving System of Equations.

When you solve an application problem that involves a system of equations we typically go through the

following steps.

Step 1: Read the question carefully and identify the x and y variables – this will typically be the two

unknowns you are being asked to find in the problem.

Step 2: Read through the question and identify two pieces of information that connect x and y as these

will be used to generate our two linear equations.

Step 3: Write down the two equations and solve the system of equations either by using the

Substitution Method or the Addition Method.

Step 4: Check your answer to see if it is correct and that it makes sense in the context of the question.

Example 1: Two numbers when added together come to 54 but when they are subtracted you get 36.

What are the two numbers?

Solution: Let x = the first number (the largest number) Step 1

Let y = the second number( the smallest number)

The part of the sentence that says “Two numbers when added together come to 54” give

us the equation x + y = 54 while the part of the sentence “when they are subtracted you get

36” gives us the equation.x – y = 36

Step 2

We now have the equations:-

Equation 1: x + y = 54

Equation 1: x – y = 36

These two equations are in standard form and can be immediately added using the addition

method we get the following.


Equation 2: x – y = 36 (add)

2x = 90

x = 45

We now use x = 45 in the equation 1: x + y = 54

45 + y = 54

y = 9

So the solution to this problem is that the two numbers are 45 and 9. Step 3

If we check our solution we get 45 + 9 = 54 and 45 – 9 = 36 Step 4


Page | 21

Example 2:

Two Complimentary angles have the property that the larger angle is 15 more than twice

the smaller angle. What are the sizes of the two angles?

Solution: Let x = the first number (the smallest angle) Step 1

Let y = the second number( the largest angle)

The part of the sentence that says “Two Complimentary angles” give us the equation

x + y = 90 while the part of the sentence that says “the larger angle is 15 more than twice

the smaller angle” gives us the equation y = 2x + 15

Step 2



Equation 2: y = 2x + 15

In this situation the Substitution Method is the simplest to use.

Substitute y = 2x + 15 into equation 1: x + y = 90

x + 2x + 15 = 90

3x + 15 = 90

3x = 75

x = 25

We now use x = 25 in the equation 2:

y = 2x + 15

y = 2(25) + 15

y = 65

So the solution to this problem is that the two angles are 250 and 65

0. Step 3

If we check our solution we get 25 + 65 = 90 and that 65 = 2(25) + 15 Step 4

x

y


Page | 22

Example 3: A truck rental company charge a daily fee and a mileage fee. A person hired a truck on two

occasions. On the first journey it cost $85 to use the truck for two days and travel 100

miles . On the second journey it cost $165 to use the truck for 3 days and travel 400 miles.

How much does the truck rental company charge per day and how much per mile?

Solution: Let x = daily fee

Let y = mileage fee Step 1

The part of the sentence that says “On the first journey it cost $85 to use the truck for two

days and travel 100 miles” give us the equation 2x + 100y = 85.

The part of the sentence that says “On the second journey it cost $165 o use the truck for 3

days and travel 400 miles.” give us the equation 3x + 1400y = 165. Step 2


Equation 1: 2x + 100y = 85

Equation 2: 3x + 400y = 165


When you look at the two equations we can eliminate the y variable by multiplying

equation 1 by – 4

2x + 100y = 85 multiply by – 4 –8x – 400y = – 340

3x + 400y = 165 3x + 400y = 165 (add)

– 5x = – 175

x = 35


2(35) + 100y = 85

70 + 100y = 85

100y = 15

y = 0.15

So the hire company charge $35 per day and $0.15 per mile Step 3

If we check our solutions

Journey 1: 2days + 100 miles = 2($35) + 100($0.15) = $70 + $15 = $85

Journey 2: 3days + 400 miles = 3($35) + 400($0.15) = $105 + $60 = $165 Step 4


Page | 23

Example 4: A shopkeeper wishes to make 20 pounds of a birdseed mixture that will cost a total of

$110. He wishes to mix nut seed that costs $4 a pound with sunflower seed that costs $6 a

pound. How much of each type of seed should he use?

Solution: Let x = the amount of nut seed to use Step 1

Let y = the amount of sunflower seed to use

The part of the sentence that says “A shopkeeper wishes to make 20 pounds of a birdseed

mixture” give us the equation x + y = 20 while the part of the sentence that says “. He

wishes to mix nut seed that costs $4 a pound with sunflower seed that costs $6 a pound.”

gives us the equation 4x + 6y = 110

Step 2





When you look at the two equations we can eliminate the y variable by multiplying

equation 1 by – 6

x + y = 20 multiply by – 6 –6x – 6y = – 120

4x + 6y = 100 4x + 6y = 110 (add)

– 2x = – 10

x = 5

We now use x = 22.5 in the equation x + y = 20

5 + y = 20

y = 15 Step 3

So the shopkeeper will mix his seed with 5 pound of nut seed and 15 pounds of sunflower

seed.


5 + 15 = 20 pound of seed and 5($4) + 15($6) = $20 + $90 = $110 Step 4


Page | 24

Example 5: A chemist wishes to make 1000 ml of a 10% solution of acid.

He only has 5% and 25% solutions of acid.

How much of each strength of acid must he use.

Solution: Let x = amount (in ml) of the 5% solution to use

Let y = amount (in ml) of the 25% solution to use Step 1

The part of the sentence that says “A chemist wishes to make 1000 ml” give us the

equation x + y = 1000 while the part of the sentence that says “A chemist wishes to make

20 ml of a 10% solution of acid..” gives us the equation 0.05x + 0.25y = 0.10(1000)

Step 2



Equation 2: 0.05x + 0.25y = 100

If we rearrange equation 1 x + y = 1000

We get the following equation y = – x + 1000

We can now use the Substitution method to solve this system of equations.

Replace the y in the second equation with y = – x + 1000 to get

0.05x + 0.25y = 100

0.05x + 0.25(– x + 1000) = 100

0.05x – 0.25x + 250 = 100

– 0.2x + 250 = 100

– 0.2x = – 150

x = 750

We now use x = 750 in the equation x + y = 1000

750 + y = 1000

y = 250

So you must mix 750 ml of the 5% solution with 250 ml of the 25% solution. Step 3


750 ml + 250 ml = 1000 ml of acid

750(0.05) + 250(0.25) = 37.5 + 62.5 = 100 ml pure acid

Which is diluted in a 1000 ml solution that gives

= 10% acid solution. Step 4


Page | 25

Exercise 3

1. Two numbers when added together come to 89 but when they are subtracted you get 49.

What are the two numbers?

2.

Two Supplementary angles have the property that the larger angle is 60 more than three

times the smaller angle. What are the sizes of the two angles?

3. A truck rental company charge a daily fee and a mileage fee. A person hired a truck on two

occasions. On the first journey it cost $140 to use the truck for four days and travel 200 miles .

On the second journey it cost $500 to use the truck for 5 days and travel 1000 miles.

How much does the truck rental company charge per day and how much per mile?

4. A food stand charges $10 for 5 hamburgers and 5 fries, while it charges $5.60 for 2 hamburgers and

fries. How much does it cost to buy a hamburger and a fries?

5. A plumber charges a fixed charge for coming to your hous plus an hourly rate to repair the leak.

One leak that took 8 hours to repair cost a total of $480 while a second leak that took 2 hours to

repair cost a total of $180. How much does the plumber charge as a fixed charge and how much

does he charge per hour for a repair?

6. A shopkeeper wishes to make 10 pounds of a coffee mixture that will cost a total of $54. He wishes

to mix Scottish Coffee that costs $15 a pound with English Coffee that costs $3 a pound. How much

of each type of coffee should he use?

7. A chemist wishes to make 500 ml of a 6% solution of acid.

He only has 5% and 10% solutions of acid.

How much of each strength of acid must he use?

x y


Page | 26

Solutions:

Exercise 1: 1. (a) Consistent 1 solution (b) Consistent 1 solution

(c) Dependant solutions (d) Inconsistent 0 solutions

(e) Consistent 1 solution

2. (a) (1,2) (b) (4, – 2) (c) no solutions (d) solutions (e) (2, – 4)

Exercise 2A: 1. (1,3) 2. (1,1) 3.(10,– 2) 4. solutions

5. 0 6. (2, – 2) 7. (10,2) 8. solutions

9. 0 solutions 10. (0, – 2 ) 11.(1,2) 12.( ½, ¼)

Exercise 2B: 1. (1,3) 2. (1,1) 3.(10,– 2) 4. solutions

5. 0 6. (2, – 2) 7. (10,2) 8. solutions

9. 0 solutions 10. (0, – 2 ) 11.(1,2) 12.( ½, ¼)

Exercise 3: 1. x = largest number = 69 y = smallest number = 20

2. x = smallest angle = 750 y = largest angle = 285

0

3. x = daily fee = $20 y = mileage fee = $0.30

4. x = cost of a hamburger = $1.20 y = cost of fries = $0.80

5.x = fixed charge = $80 y = cost per hour = $50

6.x = amount of Scottish Coffee = 2 pounds y = amount of English Coffee = 8 pounds

7.x = amount of 5% acid = 400 ml y = amount of 10% acid = 100 ml

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