olympic college topic 15 solving system of linear equations topic 15 –solving system of linear...

Olympic College Topic 15 Solving System of Linear Equations

Topic 15 –Solving System of Linear EquationsIntroduction: 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 commonformats 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 they-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 iscalled 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 satisfiesboth equations. When you try to solve a system of equations there are three distinctpossibilities.

The diagrams below show visually the three possibilities.

x

y

Situation 1:

In the above situation thetwo lines cross at one point.

In this situation we say thatthe system is Consistent

and has one solution.

x

y

Situation 2:

In the above situation the twolines are parallel and so nevercross. In this situation we saythat the system is Inconsistentand has no solutions.

x

y

Situation 3:

In the above situation thetwo lines are identical andIn this situation we say thatthe system is DependantHas a infinite solutions.

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It is possible to find the number of solutions to a system of equations by expressing each linear equationin 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 nosolutions and is called “Inconsistent”.

If both lines are identical then there are an infinite number of solutions and the system ofequations is called “Dependant”.

Example 1: For the following system of equations state the type of system of equation we have and thenumber 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 = 4y = – 2x + 42y =

y

=

– x+6– ½x+3

These equations have different slopes so we have a Consistent system of equations withone solution,.

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

y==

¼x + 4¼x – 4

– 2yy==

– ½x – 6¼x+3

These equations have the same slope but different y-intercepts so we have an Inconsistentsystem of equations no solutions.

Solution (c) 6x + 3y = 12 y = – 2x + 43y = – 6x + 12y = – 2x + 4

These equations are identical so we have a Dependant system of equations with an infinitenumber 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 readoff the point of intersection if it exists.

Example 1: Solve the system of equations yy

==

2x + 43x – 6

Solution: For the line y = 2x + 4

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

y = 2(0) + 4y = 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 solvey = – 3(0) – 6y = – 6 So the line cuts the y-axis at – 6.

To find the x-intercept, make y = 0 and solvey06

===

– 3x – 6– 3x – 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 issaid to be “Consistent” and the solution to the system of equations is the point(– 2 , 0) orit can also be expressed as x = – 2 and y = 0.

x

y

0

y = 2x+4

y = – 3x – 6


Example 2: Solve the syatem of equations y = 2x + 42y – 4x = 4

For the line y = 2x + 4

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

So the line cuts the y-axis at 4.

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

2xx

So the line cuts the x-axis at – 2.

For the line 2y – 4x = 4

To find the x-intercept, make y = 0 and solve2(0) – 4x– 4x

x

===

44

– 1So the line cuts the x-axis at – 1

To find the y-intercept, make x = 0 and solve2y – 4(0)2y

y

===

442

x0

So the line cuts the y-axis at 2

y

y = 2x+4

2y – 4x = 4

The two lines are parallel and so will never meet, this is interpreted as there is no solutionto the system of linear equations and in these situations we say that the system of equationsis “Inconsistent”.

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Example 3: Solve the syatem of equations y = 2x + 42y – 4x = 8

For the line y = 2x + 4

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

So the line cuts the y-axis at 4.

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

2xx

So the line cuts the x-axis at – 2.

For the line 2y – 4x = 8

To find the x-intercept, make y = 0 and solve2(0) – 4x– 4x

x

===

88

– 2So the line cuts the x-axis at – 2

To find the y-intercept, make x = 0 and solve2y – 4(0)2y

y

===

884

The two lines are identical and so every point on the line is a potential solution, this isinterpreted as there are an infinite number of solutions to the system of linear equationsand in these situations we say that the system of equations is “Dependant”.

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x

So the line cuts the y-axis at 4

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 numberof 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 =

(d) 4x – 3y =

14 and

24 and

4x +14y=

8x – 6y =

10

48

(e) 2x + y = 0 and 3y = 2x – 14


2. Solving a System of Equations AlgebraicallyThere are two common algebraic methods used to solve a system of equations. They are:-

A. The Substitution method

B. The Addition MethodA.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 = … orx = ….. and then substituting this new equation into the second equation. The result of doing this is thatwe will eliminate one of the two variables x or y and will be left with a linear equation in only onevariable 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 – 13Equation 2: – 4x – 7 = 9y

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

– 4x – 7– 4x – 7– 4x – 7– 4x– 22x

x

======

9y9(2x – 13)18x – 11718x – 110– 1105

We now substitute the x = 5 into equation 1:

yyy

===

2x – 132(5) – 13

– 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 = 76x – y = 1

Solution. Equation 1: 2x + 3y = 7Equation 2: 6x – y = 1

We will use equation 1 and rearrange it into the form y = …….Re-arrange 6x – y

– yy

===

1– 6x + 16x – 1

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

2x + 3y2x + 3(6x – 1)

2x + 18x – 320x – 320x

x

======

7777100.5

We now substitute the x = 0.5 into equation 2: yyy

===

6x – 16(0.5) – 12

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

Example 3. Solve the system of equations y = – 2x + 56x + 3y = 0

Solution. Equation 1: y = – 2x + 5Equation 2: 6x + 3y = 0

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

x + 3y6x + 3(– 2x + 5 )

6x – 6x + 1515

====

0000

This is clearly impossible, we interpret any situation like this where all the variablesdisappear and we are left with an impossible statement as an Inconsistent system and inthese there will be no solutions.

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Example 4. Solve the system of equations 3y + x = 103x + 4y = 15

Solution. Equation 1: 3y + x = 10Equation 2: 3x + 4y = 15

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

3y + xx

==

10– 3y + 10

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

3x + 4y3(– 3y + 10) + 4y

– 9y + 30 + 4y– 5y + 30

====

15151515

– 5y = – 15y = 3

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

==

– 3(3) + 101

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

Example 5: Solve the system of equations 2x + y = 33y = 9 – 6x

Solution : Equation 1: 2x + y = 3Equation 2: 3y = 9 – 6x

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

3y = 9 – 6xy = 3 – 2x

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

2x + y = 32x + 3 – 2x = 3

3 = 3

When we get a situation like this where the variables all disappear and we are left with atrue statement we interpret such situations as a Dependant system of equations and in thesesituations we will have solutions.

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Example 6 Solve the system of equations x = y+43x + 7y = – 18

Solution : Equation 1: 4x – 3y =Equation 2: 2x + 4y =

Re-arrange 2x + 4y =2x =

65

5– 4y + 5

x= – 2y +5

2

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

2x + 3y2x + 3(6x – 1)

2x + 18x – 320x – 320x

x

======

7777100.5

2We can the substitute the x = 0.5 into equation 2: y = 6x – 1 = 6(0.5) – 1 =The solution to the system of equations is the point(0.5, 2).

Example 7 Solve the system of equations 0.1x + 0.02y =0.2x + 0.01y

=

0.91.2

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 decimalplaces the coefficients have. In this case it is two decimal places so we will multiply theequations by 100.

Equation 1: 0.1x + 0.02y =Equation 2: 0.2x + 0.01y =

0.91.2

x 100x 100

10x + 2y20x + y

==

90120

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

We now substitute y = – 20x + 120 into the equation 10x + 2y10x + 2(– 20x + 120)

10x – 40x + 240– 30x + 240

– 30xx

======

90909090

– 1505

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

=

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

x 24

x 10

=

=

– 39

– 28

Equation 1:

Equation 2:

Rearrange

x+ y =

x+ y =

5x + 4y = – 28

8x + 18y

5x + 4y

to get4y = – 5x – 28

y =

We now substitute y = into the equation

8x + 18(

8x –

8x + 18y

)

x – 126

=

=

=

– 39

– 39

– 39

– x – 126 = – 39

– x

x

=

=

87

– 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 2ASolve the following system of equations using the Substitution Method.

1. 2x + 3y = 11 2. 4x – 4y = 04x – 2y = – 2 y = 1

3. x + 5y = 0 4. 2x – 5y = 103x – 4y = 38 6x – 15y = 30

5. x + 3y = – 2 6. x – 4y = 105x + 15y = 0 3x – 2y = 10

7. 4a + 7b = 54 8. 2x – 3y = 12a – 3b = 14 8x – 12y = 4

9. 3x + 4y = 12 10. 2a – 5b = 106x + 8y = 24 3a – b = 2

12.11. 0.3x + 0.4y =

0.02x + 0.01y =

1.1

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 samenumber of either x or y’s but one must also be negative. Once we have achieved this goal we can thenadd the two equations together and so eliminate this variable. We will then be left with a linearequation in only one variable that we can then solve in the usual way.

Example 1: Solve the system of equations x – 2yx + 2y

==

68

Solution: Equation 1: x – 2yEquation 2: x + 2y

==

68

These two equations are already in the correct form as they have a +2y and the other – 2yWe now add he two equations together and eliminate the y variable.

14

x – 2yx + 2y

2xx

====

68

7

We now know that x = 7 to find the corresponding value of y we substitute x = 7 into theequation x + 2y = 8 to get

x + 2y7 + 2y

2yy

====

881½

So the solution is (7, ½)

Note: You can always check that your solution is correct by substituting it into both equations andchecking that they work.

Using x = 7 and y = ½ in Equation 1 we get x – 2y = 67 – 2(½) = 67 – 1 = 6

(checks)6

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

=

=

6

87 + 2(½) = 87+1 = 8

8 = 8 (checks)


Example 2: Solve the system of equations 2x – 3y = 42x + y = – 4

Solution: Equation 1: 2x – 3y = 4Equation 2: 2x + y = – 4

In this example we have to change one of the equations we can either change equation 2 bymultiplying it by – 1 and by doing this we get the – x term that we need. On the other handwe could multiply equation 2 by 3 in order to get the + 3y term that we need. In thissituation we will multiply equation 2 by 3.

2x – 3y = 4 2x – 3y = 42x + y = – 4 multiply by 3 6x + 3y = – 12 (add)

8xx

==

– 8– 1

Use x = – 1 in the equation 2x + y2(– 1) + y

– 2+yy

====

– 4– 4– 4– 2

So the Solution is (– 1, – 2)

Example 3: Solve the system of equations x – y = 23x – 3y = 0

Solution: Equation 1: x – y = 2Equation 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 = – 63x – 3y = 0 3x – 3y = 0 (add)

0 = – 6

In this situation we have added the equations together and all the variables have beeneliminated. 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 nosolutions.

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x – 3y–3x

==

11– 5y – 17

Example 4: Solve the system of equations

Equation 1: 2x – 3y =

Solution: 11

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 1: 2x – 3y =Equation 2: –3x +5y =

11– 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 bothequations 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 equation2x– 3y2(4) – 3y

8 – 3y– 3y

y

=====

1111113– 1

So the solution is (4, – 1)

Example 5: Solve the system of equations 2x – 3y =– 4x + 6y =

3– 6

Solution: Equation 1: 2x – 3y =Equation 2: – 4x + 6y =

3–6

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

2x – 3y = 3 multiply by 2 4x – 6y = 64x + 5y = 6 (add)4x + 6y = – 6

0

= 0

In this situation we have added the equations together and all the variables have beeneliminated. The result is that we are left with the true statement that 0 = 0. We interpret thissituation as we have a Dependant system of equations and that there will besolutions.

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Example 6: Solve the system of equations 5x = 2y – 43x – 5y = 6

Solution: Equation 1: 5x = 2y – 4Equation 2: 3x – 5y = 6

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 =Equation 2: 3x – 5y =

– 46

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 bothequations 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 = 203x – 5y = 6 multiply by 2 (add)6x – 10y = 12

– 19x

= 32

32

x =

19

32

19We now use x = in the equation

3(

3x – 5y

32) – 5y

19

=

=

6

6

96

19 – 5y

– 5y

=

=

6

6+96

19

– 5y =210

19

y =42

19

32

19So the Solution is (

42

19, ).


Example 7 Solve the system of equations 0.2x + 0.12y =0.1x – 2.4y =

0.647.7

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 decimalplaces the coefficients have in each equation. In this case it is two decimal places forequation 1 and 1 decimal place for equation 2.

Equation 1: 0.2x + 0.12y =Equation 2: 0.1x – 2.4y =

0.647.7

x 100x 10

20x + 12yx – 24y

==

6477

We can now focus in trying to eliminate one of the variables by matching them up.In this question if we focus on the y terms we have + 12y and – 24y we can change bothinto + 12y and – 24y by multiplying equation 1 by 2.


20x + 12y = 64 multiply by2 40x + 24y = 128x – 24y = 77 x – 24y = 77 (add)

41xx

==

2055

We now use x = 5 in the equation20x + 12y20(5) + 12y100 + 12y

12yy

=====

646464

– 36– 3

So the Solution is(5, – 3)

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Example 8 Solve the system of equations x y =

x+ y =

Solution:First we change the equations so that all the coefficients are whole numbers.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 = x8 5x + 6y = 11

We can now focus in trying to eliminate one of the variables by matching them up.In this question he equations are all ready to be added and we need to do no more.


5x – 6y = – 1(add)5x + 6y

10xx

===

11101

We now use x = 1 in the equation5x + 6y5(1) + 6y

5 + 6y6yy

=====

111111

61

So the Solution is(1, 1)

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Exercise 2BSolve the following system of equations using the Addition Method.

1. x + 3y = 10 2. x – y = 03x – 3y = – 6 2x – 3y = – 1

3.

5.

7.

2x + 5y =3x – 4y

=

2x + 5y =10x + 25y =

4a + 7b =

3022

– 50

54

4.

6.

8.

6x – 15y =2x – 5y =

3x – 4y =2x – 2y =

2x – 3y =

3010

148

12a – 3b = 14 8x – 12y = 4

9. 3x + 4y = 12 10. 2a – 3b = 66x + 8y = 24 3a – 2b = 4

11. 0.6x + 0.8y = 2.2 12. =

0.01x + 0.005y = 0.02 =


3. Application Problems involving System of Equations.

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

Step 1: Read the question carefully and identify the x and y variables – this will typically be the twounknowns 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 thesewill be used to generate our two linear equations.

Step 3: Write down the two equations and solve the system of equations either by using theSubstitution 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 1Let y = the second number( the smallest number)

The part of the sentence that says “Two numbers when added together come to 54” giveus the equation x + y = 54 while the part of the sentence “when they are subtracted you get36” gives us the equation.x – y = 36

Step 2We now have the equations:-

Equation 1:Equation 1:

x+yx – y

==

5436

These two equations are in standard form and can be immediately added using the additionmethod we get the following.

Equation 1: x+y = 54Equation 2: (add)x – y

2xx

===

369045

We now use x = 45 in the equation 1: x+y45 + y

y

===

54549

So the solution to this problem is that the two numbers are 45 and 9.If we check our solution we get 45 + 9 = 54 and 45 – 9 = 36

Step 3Step 4

Page | 20

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

The part of the sentence that says “Two Complimentary angles” give us the equationx + y = 90 while the part of the sentence that says “the larger angle is 15 more than twicethe smaller angle” gives us the equation y = 2x + 15



x+yy

==

902x + 15

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

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

3x + 153x

x

=====

9090907525

We now use x = 25 in the equation 2:

yyy

===

2x + 152(25) + 1565

So the solution to this problem is that the two angles are 250 and 650.

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

Step 3

Step 4

Page | 21


Example 2:

y

x

Two Complimentary angles have the property that the larger angle is 15 more than twicethe smaller angle. What are the sizes of the two angles?


Example 3: A truck rental company charge a daily fee and a mileage fee. A person hired a truck on twooccasions. 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 feeLet y = mileage fee Step 1

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

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

We now have the equations:-


2x + 100y3x + 400y

==

85165

We can now focus in trying to eliminate one of the variables by matching them up.When you look at the two equations we can eliminate the y variable by multiplyingequation 1 by – 4

2x + 100y = 85 multiply by – 4 –8x – 400y = – 3403x + 400y = 165 3x + 400y = 165 (add)

– 5xx

==

– 17535

We now use x = 35 in the equation 2x + 100y = 852(35) + 100y = 85

70 + 100y100y

y

===

85150.15

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

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 = $165Step 4

Page | 22


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

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

The part of the sentence that says “A shopkeeper wishes to make 20 pounds of a birdseedmixture” give us the equation x + y = 20 while the part of the sentence that says “. Hewishes to mix nut seed that costs $4 a pound with sunflower seed that costs $6 a pound.”gives us the equation 4x + 6y = 110


Equation 1: x+y = 20Equation 2: 4x + 6y = 110

We can now focus in trying to eliminate one of the variables by matching them up.When you look at the two equations we can eliminate the y variable by multiplyingequation 1 by – 6

x+y = 20 multiply by – 6 –6x – 6y = – 1204x + 6y = 100 4x + 6y = 110 (add)

– 2xx

==

– 105

We now use x = 22.5 in the equationx+y5+y

==

2020

y = 15 Step 3

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


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

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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 useLet 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 theequation x + y = 1000 while the part of the sentence that says “A chemist wishes to make20 ml of a 10% solution of acid..” gives us the equation 0.05x + 0.25y = 0.10(1000)



x+y0.05x + 0.25y

==

1000100

If we rearrange equation 1We get the following equation

x+yy

==

1000– 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 + 0.25y0.05x + 0.25(– x + 1000)

0.05x – 0.25x + 250– 0.2x + 250– 0.2x

x

======

100100100100– 150750

We now use x = 750 in the equation x+y = 1000750 + y = 1000

y = 250

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


750 ml + 250 ml = 1000 ml of acid750(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

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

3.

4.

5.

6.

7.

A truck rental company charge a daily fee and a mileage fee. A person hired a truck on twooccasions. 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?

A food stand charges $10 for 5 hamburgers and 5 fries, while it charges $5.60 for 2 hamburgers andfries. How much does it cost to buy a hamburger and a fries?

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 torepair cost a total of $180. How much does the plumber charge as a fixed charge and how muchdoes he charge per hour for a repair?

A shopkeeper wishes to make 10 pounds of a coffee mixture that will cost a total of $54. He wishesto mix Scottish Coffee that costs $15 a pound with English Coffee that costs $3 a pound. How muchof each type of coffee should he use?

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?

Page | 25

x y


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)5. 0

2. (1,1)6. (2, – 2)

3.(10,– 2)7. (10,2)

4.8.

solutionssolutions

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

Exercise 2B: 1. (1,3)5. 0

2. (1,1)6. (2, – 2)

3.(10,– 2)7. (10,2)

4.8.

solutionssolutions

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

Exercise 3: 1. x = largest number = 692. x = smallest angle = 750

3. x = daily fee = $204. x = cost of a hamburger = $1.205.x = fixed charge = $806.x = amount of Scottish Coffee = 2 pounds7.x = amount of 5% acid = 400 ml

y = smallest number = 20y = largest angle = 2850

y = mileage fee = $0.30y = cost of fries = $0.80y = cost per hour = $50y = amount of English Coffee = 8 poundsy = amount of 10% acid = 100 ml

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olympic college topic 15 solving system of linear equations topic 15 –solving system of linear...

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