Substitution and Elimination

 

These are both methods of solving a system of equations with variables, for example:

 

x + 2y = 7

2x + y = 8

 

The point of substitution and elimination is to find the values of the variables by applying various mathematical methods to the equations so that the variable and it’s value are the only things left.  It can also be used for solving equations with three or more variables, or with three or more equations.  We will first look at substitution.

 

Substitution

 

The first method of substitution we will look at is solving two equations with two variables.  The idea of substitution is to isolate one variable and place the corresponding value in the other equation.    For example:

 

1.  3x - 6y = 9

2.  2x + 3y = 27

 

We will isolate the variable “x” in the first equation by moving the other variable (“y”) to the other side of the equation and eliminating x’s coefficient (3).  A trick for doing this is to glance at your equations and find which variable will divide most evenly into the other terms of the equation, giving you the simplest numbers to work with and avoiding awful things like decimals and fractions, which are terrible and horrible:

 

3x – 6y = 9                             -6y goes to the other side, becoming +6y

3x / 3= (9 + 6y) / 3          Both sides are divided by 3 to eliminate the 3 in front of the x

x = 3 + 2y

 

Now that we have a value for x, we can place our new x value (3 + 2y) from our first equation in the place of x in our second equation and solve for y to get its value:

 

2x + 3y = 27                                  Substitute (3 + 2y) where x is

2(3 + 2y) + 3y = 27          Multiply out the bracket to get your new values

6 + 4y + 3y = 27                    Combine the y terms

6 + 7y = 27                    Move your 6 to the other side of the equation, making it –6

7y = 27 – 6                    Subtract 6 from 27

7y / 7= 21 / 7                    Divide both sides by seven to get the y value

y = 3

 

Now that you know that y=3, substitute that value into either of the first two equations and solve for x, which will solve our equations.  We will use the first equation:

 

3x – 6y = 9                                      Substitute 3 for y

3x – 6(3) = 9                                    Multiply 6 by 3

3x – 18 = 9                    Move –18 to the other side of the equation, making it +18

3x = 9 + 18                    Add 9 and 18

3x / 3 = 27 / 3                    Divide both sides by three to get the value of x

x = 9

 

The equations have now been solved using substitution.  Our answer will read:

 

x = 9

y = 3

 

It’s a good idea to substitute both numbers into the equations to make sure they equal out:

 

1.  3(9) – 6(3) = 9

     27 – 18 = 9

     9 = 9

 

2.     2(9) + 3(3) = 27

18 + 9 = 27

27 = 27

 

Examples

 

Here are some examples of equations we will solve with substitution, progressing from easier to more difficult. 

 

Example I: These are very simple equations to solve, as the variables have no coefficients, cutting down possible complications:

 

1.  x + y = 11

2.  x – y = 3

 

We must single out one variable to substitute in the other equation, and we will use the second equation and single out x:

 

x – y = 3                       Move the –y to the other side, making it +y

x = 3 + y

 

Now substitute that value into the first equation in the place of x and solve for y:

 

x + y = 11                     Substitute (3 + y) in the place of x

(3 + y) + y = 11                              Drop your brackets

3 + y + y = 11                                  Combine your y variables

3 + 2y = 11                                      Move your 3 to the other side, becoming –3

2y = 11 – 3                                      Subtract 3 from 11

2y / 2 = 8 / 2                                    Divide both sides by 2 to get the value of y

y = 4          

 

Now that you have the value for y, substitute it into one of your first two equations.  We will use the second:

 

x – y = 3                       Substitute 4 for y

x –  4 = 3                      Move –4 to the other side, becoming +4

x = 7

 

x = 7

y = 4

 

Example II: These are regular equations, like the two used in the explanation.  They are the most common type of equations you will solve:

 

1.  4x + 3y = 20

2.  2x + 6y = 28

 

We need to get one variable alone so we can substitute, so we will use x in the second equation as it will divide out the nicest:

 

2x + 6y = 28                                  Move 6y to the other side, becoming –6y

2x / 2 = (28 – 6y) / 2                   Divide both sides by 2 to get your x value

x = 14 – 3y

 

Our new x value can now be placed in the first equation and we can solve for y:

 

4x + 3y = 20                                  Substitute (14 – 3y) in place of x

4(14 – 3y) + 3y = 20                   Multiply out your brackets with the 4

56 –12y + 3y = 20                   Combine your y variables

56 – 9y = 20                                  Move 56 to the other side, becoming -56

- 9y = 20 – 56                                  Subtract 56 from 20

- 9y / -9= - 36 / -9                           Divide both sides by –9 to get the y value

y = 4

 

Now place substitute your y value in one of the first two equations and solve for x.  We will use the second equation:

 

2x + 6y = 28                                  Substitute 4 for y

2x + 6(4) = 28                                  Multiply 6 by 4

2x + 24 = 28                                  Move 24 to the other side, becoming –24

2x = 28 – 24                                  Subtract 24 from 28

2x / 2 = 4 / 2                                    Divide both sides by 2 to get the x value

x = 2

 

x = 2

y = 4

 

 

 

 

Example III: These equations have fractions in them, so it will be a little harder due to the complications coming from combining them with substitution, making things a bit confusing:

 

1.     6x + 2y = 2 4/5

2.     4x + 3y = 2 8/15

 

We must first, of course, centralize one variable.  We will use y in the first equation:

 

6x + 2y = 2 4/5                                 Move 6x to the other side, becoming –6x

2y /2 = (2 4/5 – 6x) / 2                   Divide both sides by 2 to get the y value

y = 1 2/5 – 3x

 

Now substitute the new y value in the second equation and solve for x:

 

4x + 3y = 2 8/15                                                 Substitute (1 2/5 – 3x) for y

4x + 3(1 2/5 – 3x) = 2 8/15                   Multiply out the brackets by 3

4x + 4 1/5 – 9x = 2 8/15                   Combine the y variables

-5x + 4 1/5 = 2 8/15          Move 4 1/5 to the other side, becoming –4 1/5

-5x = 2 8/15 – 4 1/5          Find a common denominator for 2 8/15 and 4 1/5

-5x = 2 8/15 – 4 3/15          Subtract the two terms

-5x / -5 = -1 2/3 / -5          Divide both sides by -5 to get the value of x

x = 1/3

 

Place that value into one of the first two equations and solve for y.  We will use the first:

 

6x + 2y = 2 4/5                                                    Substitute 1/3 for x

6(1/3) + 2y = 2 4/5                          Multiply 6 by 1/3

2 + 2y = 2 4/5                                                    Move 2 to the other side, becoming –2

2y = 2 4/5 – 2                                                    Combine the two terms

2y / 2 = 4/5 / 2                                                     Divide both sides by 2

y = 2/5

 

x = 1/3

y = 2/5

 

Example IV: A regular set of equations, with a twist in the answer.  Can you guess what it is?  Here’s a hint - look at the third term of the second equation:

 

1.     5x + 5y = 7

2.     10x – 6y = 4.4

 

First get one variable alone.  We will use x in the first equation:

 

5x + 5y = 7                                      Move 5y to the other side, becoming –5y

5x / 5 = (7 – 5y) / 5                   Divide both sides by 5 to get the x value

x = 1.4 –y

 

Now substitute the new x value into the second equation:

 

10x – 6y = 4.4                                 Substitute (1.4 – y) for x

10(1.4 – y) – 6y = 4.4                   Multiply out the brackets by 10

14 – 10y – 6y = 4.4                   Combine the y variables

14 – 16y = 4.4                                 Move 14 to the other side, becoming –14

-16y = 4.4 – 14                              Combine the two terms

-16y / -16 = -9.6 / -16                   Divide both sides by –16 to get the y value

y = 0.6

 

Now substitute your new value into one of the first two equations.  We will use the first:

 

5x + 5y = 7                                      Substitute 0.6 for y

5x + 5(0.6) = 7                                 Multiply 5 by 0.6

5x + 3 = 7                     Move 3 to the other side, becoming –3

5x = 7 – 3                    Combine the two terms

5x / 5 = 4 / 5                                    Divide both sides by 5 to get the x value

x = 0.8                          

 

x = 0.8

y = 0.6

 

 

 

 

 

 

Practice Questions

 

Here are seven questions to try on your own.  The answers are in the appendix.  The last is a word question that requires you to pick out the information and make your own equation.

 

1.  x + y = 12

     x – y = 4

 

2.  2x + 4y = 26

     4x + 2y = 34

 

3.  2x + 3y = 23

     3x – 4y = -8

 

4.  1/3x + 2y = 1 2/3

4x + 2/3y = 16 1/9

 

5.     1.4x – 3y = -5.6

3x + 2.8y = 6.8

 

6.    7x + 3y = 3.425

6x – y = - 0.1

 

7.    John went to the store and bought 6 jars of pickles and 3 bottles of ketchup for $12.00.  The next week he bought 4 jars of pickles and 5 bottles of ketchup for $12.50.  How much does each product cost?

 

Elimination

 

Elimination is also used to solve for variables in equations, but involves (obviously) eliminating instead of substituting.  The general idea is to combine the equations so that they eliminate one variable, allowing you to solve for the other.  Our example equations will be:

 

1.     2x + 2y = 14

2.     4x – 2y =  4

 

The first step is to combine the equations.  This is done by combining the like terms into a new “super-equation” if you will.  Ideally, one of the variables will have coefficients that will cancel each other out when added (ie. 5 and –5 will equal 0).  If not, you need to play around with it a bit, but that will be explained later.  Luckily, this equation is set up quite lovely and the y variables will cancel.  So we combine:

 

   2x + 2y = 14

+ 4x – 2y = 4

6x + 0 = 18 or 6x = 18                    

 

The y variables leave an equation that we can easily solve:

 

6x / 6 = 18 / 6                                  Divide both sides by 6 to get the x value

x = 3

 

I must say this is a heck of a lot easier than substitution.  Of course, only for certain equations.  Complex equations will require a lot of work to combine as will come up later.  Anyway, place your x value in one of your first two equations and solve for y.  We will use the first:

 

2x + 2y = 14                                  Substitute x for 3

2(3) + 2y = 14                                  Multiply 2 by 3

6 + 2y = 14                                      Move 6 to the other side, becoming –6

2y = 14 – 6                                      Combine the terms

2y / 2 = 8 / 2                                    Divide both sides by 2 to get the y value

y = 4

 

Remarkable eh?  We should, to make sure we are absolutely correct, check our values by putting them into the original equations:

 

1.     2x + 2y = 14

2(3) + 2(4) = 14

6 + 8 = 14

14 = 14

 

2.     4x – 2y = 4

4(3) – 2(4) = 4

12 – 8 = 4

4 = 4

 

Examples

 

Now are some examples of equations we will solve with elimination, each increasing in difficulty and adding new “twists” to the process.

 

Example I: These are fairly simple equations that require a bit of tinkering before we can eliminate:

 

1.     3x + 4y = 27

2.     5x + 4y = 37

 

We can’t eliminate yet because our variable coefficients will not cancel each other out.  But luckily the y coefficients are the same value, so we can multiply one equation by –1, allowing us to successfully eliminate the y variable.  We will use the first equation, as the values are smaller in the other terms and when added will leave us nice positive numbers. 

 

(3x + 4y = 27)(-1)          Multiply the equation by –1

- 3x – 4y = - 27

 

We can combine the equations and eliminate the y variables

-3x – 4y = - 27

+ 5x + 4y = 37  

2x + 0 = 10 or 2x = 10

 

Now we can solve for x with the “super-equation”:

 

2x / 2 = 10 / 2                                  Divide both sides by 2 to get the x value

x = 5

 

Our x value can be placed into one of the first two equations and we can solve for y.  We will use the first:

 

3x + 4y = 27                                  Substitute 5 for x

3(5) + 4y = 27                                  Multiply 3 by 5

15 + 4y = 27                                  Move 15 to the other side, becoming –15

4y = 27 – 15                                  Combine the terms

4y / 4 = 12 / 4                                  Divide both sides by 4 to get the y value

y = 3

x = 5

y = 3

 

Example II: Also simple equations requiring a different kind of manipulation:

 

1.     – 2x + 2y = - 2

2.     4x + y = 34

 

In these equations, again our initial variable coefficients won’t cancel each other out.  But there is a light at the end of the tunnel.  We have a negative, the –2x, which we can multiply by 2 so we can eliminate it with the 4x of our second equation.  This will require the multiplication of the entire first equation by 2. 

(- 2x + 2y = - 2)(2)                   Multiply the equation by 2

-4x + 4y = - 4

 

Now we can combine:

 

-4x + 4y = -4

+ 4x + y + 34

0 + 5y = 30 or 5y = 30

 

The “super-equation”!  The next logical course of action is to solve for y:

 

5y / 5 = 30 / 5                                  Divide both sides by 5 to get the value of y

y = 6

 

The y value can be placed into one of the first two equations and we can find the value of x.  We will use the second equation:

 

4x + y = 34                                      Substitute 6 for y

4x + 6 = 34                                      Move 6 to the other side, becoming – 6

4x = 34 – 6                                      Combine the terms

4x / 4 = 28 / 4                                  Divide both sides by 7 to get the value of x

x = 7

 

x=7

y=6

 

Example III: We now progress into the question that requires our astute mental prowess.  These equations will use both of the twists and manipulation of the first two examples:

 

1.     3x + 2y = 20

2.     6x + 3y = 33

 

Now what can we eliminate?  Nothing right off hand.  Can we multiply it by –1 to get something to eliminate?  No, that won’t do it.  Can we multiply one equation by some term that will allow us to eliminate?  No, that too is an unsatisfactory course of action.  But can we combine those two suggestions, and multiply by a negative term so to eliminate?  Why yes, we can.  If we multiply 3x of the first equation by –2, we can eliminate it with 6z of the second equation, like so:

 

(3x + 2y = 20)(-2)                   Multiply the equation by -2 

-6x – 2y = - 20                                 

 

And then we combine the equations:

 

-6x – 2y - = - 20

+ 6x + 3y = 33

0 + 1y = 7 or y = 7

 

To our great luck, there is no need to solve for y as our combining does so for us, so we skip that bit and place our y value into one of the first two equations.  We shall use the first:

 

3x + 2y = 20                                  Substitute 7 for y

3x + 2(7) = 20                                  Multiply 2 by 7

3x + 14 = 20                                  Move 14 to the other side, becoming –14

3x = 20 – 14                                  Combine the terms

3x / 3 = 6 / 3                                    Divide both sides by 3 to get the x value

x = 2

 

x= 2

y = 7

 

 

 

 

 

Example IV: These equations have absolutely nothing in common.  Should be fun:

 

1.     7x + 2y = 25

2.     3x + 9y = 27

 

To solve these equations we must multiply both equations by something that will allow us to eliminate, as the possibility of an easy solution seems to have eluded us this time.  So what will we use?  Well, we’ve been eliminating x the last few times first, so let’s eliminate the y.  The simple route is to multiply the equation with 2y by 9 and the equation with 9y by 2.  Also, one should be negative to allow us to eliminate and we will go with the 2 so as to keep our positive numbers when we combine, which is just easier.  You may think this is a rather random way to solve this.  Yeah.  That’s what you gotta do with no clear options:

 

(7x + 2y = 25)(9)                            Multiply the equation by 9

 63x + 18y = 225

 

(3x + 9y = 27)(-2)                   Multiply the equation by -2

-6x – 18y = - 54

 

 combine the two new equations:

 

63x + 18y = 225

+ - 6x – 18y = - 54

57x + 0 = 171 or 57x = 171

 

Now we solve for x:

 

57x / 57 = 171 / 57                   Divide both sides by 57 to get the x value

x = 3

 

Place the new x value into one of the first two equations and solve for x.  We will use the first:

 

7x + 2y = 25                                  Substitute x with 3

7(3) + 2y = 25                                  Multiply 7 by 3

21 + 2y = 25                                  Move 21 to the other side, becoming –21

2y = 25 – 21                                  Combine the terms

2y / 2 = 4 / 2                                    Divide both sides by 2 to get the y value

y = 2

 

x = 3

y = 2

 

Practice Questions

 

Here are seven questions to try using elimination.  The answers will be in the appendix:

 

1.     x + y = 12

x – y = - 4

 

2. .4x – 2y = 6

      3x + 2y = 22

 

3.      2x + 3y = 19

2x + 2y = 14

 

4.     4x – 7y = 10

5x + 14y = 58

 

5.     -3x – 2y = - 18

-4x – 7y = - 37

 

6.  3x – y = 5n

      4x + y = 9n

 

7.  Paul went to the store and bought two records and a poster for $23.99.  The next month he bought one record and 4 posters for $43.46.  What was the cost of each product?

 

 

 

 

 

 

 

Elimination With Three Variables

 

The process for eliminating with three variables is similar to that of two, but with some additional steps.  Our explanatory equations will be:

 

1.     2x + 3y + 2z = 20

2.     3x – 3y + 4z = 19

3.     4x + 3y – 2z = 10

 

The first step is to move from three equations to two.  “How?” you may ask.  Well, we must take two equations and eliminate one variable, then take two other equations (obviously one will be an equation used in the first elimination) and eliminate the same variable.  From there we pick up our old eliminating techniques we just learned.  So what will we eliminate?  Well, we have a -3y and two 3y in various equations, so we will eliminate the 3y equations with the -3y and go from there.  Funny, the luck we had with that… like it was all arranged to work out.  This sounds confusing but should make sense momentarily:

 

2x + 3y + 2z = 20

+ 3x – 3y + 4z = 19

5x + 0 + 6z = 39 or 5x + 6z = 39

 

4x + 3y – 2z = 10

+ 3x – 3y + 4z = 19

7x + 0 + 2z = 29 or 7x + 2z = 29

 

These become our new equations:

 

4.     5x + 6z = 39

5.     7x + 2z = 29

 

Now we eliminate with these to get our x and z values.  Unfortunately, there is no clear elimination, so we must play with it a bit.  It looks like if we multiply the fifth equation by –3 we can cancel the z variables.  In three variable elimination, rarely is the two variable part spelled out for you.  Anyway:

 

 

(7x + 2z = 29)(-3)                    Multiply the equation by –3

-21x – 6z = - 87

 

Now we combine with equation four:

 

5x + 6z = 39

+ -21x – 6z = - 87

-16x + 0 = - 48 or –16x = - 48

 

And solve for x:

 

- 16x / -16 = - 48                   / -16                   Divide both sides by –16 to get the x value

x = 3

 

With our new x value, we can solve for z by using equation four or five.  We’ll use four:

 

5x + 6z = 39                                  Substitute 3 for x

5(3) + 6z = 39                                  Multiply 5 by 3

15 + 6z = 39                                  Move 15 to the other side, becoming –15

6z = 39 – 15                                  Combine the terms

6z / 6 = 24 / 6                                  Divide both sides by 6 to get the z value

z = 4

 

Are we done?  No, we still have to get our y value by placing the two values we found in one of the first three equations and solving for y.  We will use the first equation:

 

2z + 3y + 2z = 20                         Substitute 3 for x

2(3) + 3y + 2z = 20                   Substitute 4 for z

2(3) + 3y + 2(4) = 20                   Multiply out both brackets

6 + 3y + 8 = 20                              Combine like terms

3y + 14 = 20                                  Move 14 to the other side, becoming –14

3y = 20 – 14                                  Combine the terms

3y / 3 = 6 / 3                                    Divide both sides by 3 to get the y value

y = 2

 

Our answer will be:

 

x = 3

y = 2

z = 4

 

Of course, we should check our answers.  We will put our values into the second and third equations to make sure it works out. a good habit to form:

 

2.  3x – 3y + 4z = 19

     3(3) – 3(2) + 4(4) = 19

     9 – 6 + 16 = 19

     19 = 19

 

3.  4x +3y – 2z = 10

     4(3) + 3(2) – 2(4) = 10

     12 + 6 – 8 = 10

     10 = 10

 

 

Examples

 

 

Example 1: Usually we start with the super easy question here, but instead it will be a regular question.  It will need some work before we can eliminate:

 

1.     x + 2y + 3z = 29

2.     2x – 3y + 4z = 25

3.     – 4x + 5y + 5z = 25

 

Now what variable should we eliminate?  It must be the same for both new equations.  X seems to be the best option.  Equation one can be multiplied by –2 to eliminate with equation two, and equation two can be doubled to eliminate the x with equation three.  We will fix equation one:

 

(x + 2y + 3z = 29)(-2)                                   Multiply the equation by –2

-2x – 4y – 6z = - 58

 

Combine with the second equation:

 

-2x – 4y – 6z = - 58

+ 2x – 3y + 4z = 25

0 – 7y – 2z = - 33 or 7y + 2z = 33

 

Now we doctor equation two:

 

(2x – 3y + 4z = 25)(2)                    Multiply the equation by 2

4x – 6y + 8z = 50

 

And eliminate with equation three:

 

- 4x + 5y + 5z = 25

+ 4x – 6y + 8z = 50

 0 – y + 13z = 75 or – y + 13z = 75

 

Our two new equations are:

 

4.  7y + 2z + 33

5.  - y + 13z = 75

 

Now this is a fairly good spot to be in.  We can eliminate by multiplying the fifth equation by 7, allowing for the cancellation of the y variable:

 

(- y + 13z = 75)(7)                            Multiply the equation by 7

- 7y + 91z = 525                                              

 

And combine with equation four:

 

- 7y + 91z = 525

+ 7y + 2z = 33

0 + 93z = 558 or 93z = 558

 

Solve for z:

 

93z / 93 = 558 / 93                   Divide both sides by 92 to get the z value

z = 6

 

Now we take our z value and place it into either our fourth or fifth equation to solve for y.  We will use the fourth:

 

7y + 2z = 33                                  Substitute 6 for z

7y + 2(6) = 33                                  Multiply 2 by 6

7y + 12 = 33                                  Move 12 to the other side, becoming –12

7y = 33 – 12                                  Combine the terms

7y / 7 = 21 / 7                                  Divide both sides by 7 to get the v value

y = 3

 

Of course, now we take our z and y values and place them into one of the first three equations and solve for x.  We will use the first:

 

x + 2y + 3z = 29                              Substitute 3 for y

x + 2(3) + 3z = 29                   Substitute 6 for z

x + 2(3) + 3(6) = 29                   Multiply out the brackets

x + 6 + 18 = 29                              Combine like terms

x + 24 = 29                                      Move 24 to the other side, becoming –24

x = 29 – 24                                      Combine the terms

x = 5

 

And the answer is:

 

x = 5    y=3    z=6

Example 2:.  These equations will have nothing in common and more than likely result in large numbers as we try to solve:

 

1.     12x – 5y + 6z = 13

2.     14x + 8y – 5z =  5

3.     18x – 3y – 4z = - 2

 

What do we have to eliminate?  Nothing.  Anything we can double or turn into a negative?  Nope.  So what can we do?  The tried and true method of trial and error.  What variable will we eliminate?  Let’s say z as it has yet to be eliminated at this stage in our previous questions.  Z actually has a half-decent set up.  We can multiply the 6z by 5 and the –5z by 6 and eliminate there.  Also, we can multiply the 6z by 4 and the –4z by 6 and eliminate there:

 

(12x – 5y + 6z = 13)(5)                    Multiply the equation by 5

60x – 25y + 30z = 65

 

(14x + 8y – 5z = 5)(6)                      Multiply the equation by 6

84x + 48y – 30z = 30

 

60x – 25y + 30z = 65

+ 84x + 48y – 30x = 30

144x + 23y + 0 = 95 or 144x + 23y = 95

 

And get the other equation:

 

(12x – 5y + 6z = 13)(4)                    Multiply the equation by 4

48x – 20y + 24z = 52

 

(18x – 3y – 4z = - 2)(6)                    Multiply the equation by 6

108x – 18y – 24z = - 12              

 

108x – 18y – 24z = - 12

+ 48x – 20y + 24z = 52

156x – 38y + 0 = 40 or 156z – 38y = 40

 

Our two new equations are:

 

4.     144x + 23y = 95

5.     156x – 38y = 40

 

Bad setup.  We will eliminate the y variables now, as they are lower and will keep our numbers smaller and easier to work with.  We’ll multiply equation four by 38 and equation five by 23, so as to be able to cancel the variables:

 

(144x + 23y = 95)(38)                          Multiply the equation by 38

5472x + 874y = 3610

 

(156x – 38y = 40)(23)                          Multiply the equation by 23

3588x - 874y = 920

 

And we eliminate:

 

5472x + 874y = 3610

+ 3588x - 874y = 920

9060x – 0 = 4530 or 9060x = 4530

 

Now solve for x:

 

9060x / 9060 = 4530 / 9060          Divide both sides by 9060 to get the x value

x = 0.5

 

And now we place that into equation four and five and solve for y.  We’ll use number four:

 

144x + 23y = 95                              Substitute 0.5 for x

144(0.5) + 23y = 95                   Multiply 144 by 0.5

72 + 23y = 95                                  Move 72 to the other side, becoming –72

23y = 95 – 72                                  Combine the terms

23y / 23 = 23 / 23                   Divide both sides by 23 to get the y value

y = 1

 

This stuff takes forever.  Be very careful when doing your math… one sign mixed up and the whole thing is shot.  Concentration is the word of the day.  Now to solve for z with one of the first three equations.  We’ll use the first:

 

12x – 5y + 6z = 13                         Substitute 0.5 for x

12(0.5) – 5y + 6z = 13                                  Substitute 1 for y

12(0.5) – 5(1) + 6z = 13                   Multiply out the brackets

6 – 5 + 6z = 13                                                 Combine like terms

6z + 1 = 13                                                         Move 1 to the other side, becoming –1

6z = 13 – 1                                                          Combine the terms

6z / 6 = 12 / 6                                                     Divide both sides by 6 to get the z value

z = 2

 

Our answer is:

 

x = 0.5

y = 1

z = 2

 

Practice Questions

 

Here are three questions for you to try on elimination with three variables.  Enjoy!

 

1.     2x + 3y + 4z = 29

3x + 2y – 4z = - 4

5x + 4y – 4z = 6

 

2.     x + y + 2z = 15

4x – 2y + z = 22

3x – 4y + 5z = 15

 

3.     4x + 5y + 8z = 301

3x + 2y – z = 109

5x – 9y + 5z = 12

 

Closing Notes

 

When dealing with substitution and elimination, there are a few key points to remember for both.  First, questions can be hidden in many forms.  They can appear like our word questions, but are also sneaked into questions involving phone plans or coin amounts.  The key when reading a question is to see if you can get similar variables and a value for it to equal.  Then you’re on your way.  Another thing is that when looking at your equations, for substitution, on that first step try to use the equation that divides easiest.  It makes the rest so much easier.  For elimination, when multiplying try to keep the numbers positive and as small as possible, also for ease of use.  And avoid fractions or decimals… they were created solely to confuse you.  But the key thing is to pay attention when you do these questions.  One little mistake will mess up everything else (and I speak from experience making this kit).  And always check your values at the end, even if just in your head.  Always. 

 

Quiz

 

You’ve made it through the tutorial part, now it’s time to test your newfound skills in the quiz.  Good luck!

 

1.     Solve the following with substitution:

 

a.     2x + 3y = 17

x + 2y = 10

 

b.     3x + 6y = 36

4x + 2y = 30

 

c.     2x + 4y = - 6

3x – 2y = 23

 

d.     1/3x + 3y = 28

2x + 2/3y = 12

 

2.     Solve the following using elimination:

 

a.  5x + 3y = 25

4x – 3y = -7

 

b.     2x + 5y = 46

2x – 4y = -26

 

c.      5x – 3y = 26

3x + 6y = 0

 

d.     9x + 8y = 9.5

5x + 6y = 5.9

 

e.     4x + 2y – z = 17

3x – 2y + 4z = 6

x – 2y + 9z = 17

 

f.   2x – 4y + 3z = - 29

4x – 6y + 5z = - 41

5x – 8y + 7z = - 55

 

 

 

3.     George goes to the store one day and buys a box of Cheerios and two cartons of milk for $6.50.  The next week he buys two boxes of Cheerios and one carton of milk for $7.00.  How much does each item cost?

 

4.     Herman, desiring a hipper look, goes to trendy clothing store and buys three pairs of bellbottoms and a Nehru jacket for $65.  His friend Gerry, seeing how cool Herman now looks, goes to the same store and buys two pairs of bellbottoms and two Nehru jackets for $70.  How much does each item cost?

 

5.     Johnny, having little luck with the ladies, decides to change his hairstyle and odor so as to impress them.  He buys two bottles of shampoo, three bottles of hair gel and one bottle of cologne for $24.  The next month, having success with his new endeavor, he buys one bottle of shampoo, two bottles of hair gel, and two bottles of cologne for $24.  The next month his results are so good that he buys two more bottles of shampoo, three of hair gel, but no cologne as he had some left over, for $17.  How much does each product cost?

 

Appendix

 

Substitution:

 

1.  x + y = 12

     x – y = 4 = x = y + 4

 

(y + 4) + y = 12

y + 4 + y = 12

2y = 8

y = 4

 

x – 4 = 4

x = 8

 

2.  2x + 4y = 26 = x = - 2y + 13

     4x + 2y = 34

 

4(-2y + 13) + 2y = 34

-8y + 52 + 2y = 34

-6y = -18

y = 3

2x + 4(3) = 26

2x + 12 = 26

2x = 14

x = 7

 

3.  2x + 3y = 23

     3x – 4y = -8 = y = 0.75x + 2

 

2x + 3(0.75x +2) = 23

2x + 2.25x + 6 = 23

4.25x = 17

x = 4

 

3(4) – 4y = -8

12 – 4y = -8

-4y = -20

y = 5

 

4.  1/3x + 2y = 1 2/3 = x = - 6y + 5

4x + 2/3y = 16 1/9

 

4(-6y +5) + 2/3y = 16 1/9

-24y + 20 + 2/3y = 16 1/9

-23 1/3y = - 3 8/9

y = 1/6

 

1/3x + 2(1/6) = 1 2/3

1/3x + 1/3 = 1 2/3

1/3x = 1 1/3

x = 4

 

5.     x – 3y = -5.6 = x = 3y – 5.6

3x + 2.8y = 6.8

 

3(3y – 5.6) + 2.8y = 6.8

9y – 16.8 + 2.8y = 6.8

11.8y = 23.6

y = 2

 

x – 3(2) = -5.6

x – 6 = -5.6

x = 0.4

 

6.    7x + 3y = 3.425

6x – y = - 0.1 = y = 6x + 0.1

 

7x + 3(6x + 0.1) = 3.425

7x + 18x + 0.3 = 3.425

25x = 3.125

x = 0.125

 

6(0.125) – y = - 0.1

0.75 – y = - 0.1

y = 0.85

 

7.    6p + 3k = 12 = k = - 2p + 4

     4p + 5k = 12.50

 

4p + 5(-2p + 4) = 12.5

4p – 10p + 20 = 12.5

-6p = - 7.5

p = 1.25

 

6(1.25) + 3k = 12

7.5  + 3k = 12

     3k = 4.5

     k = 1.5

 

Elimination:

 

1.     x + y = 12

x – y = - 4

 

x + y = 12

+ x – y = -4

2x – 0 = 8

 

2x = 8

x = 4

 

4 + y = 12

y = 8

 

2.   4x – 2y = 6

      3x + 2y = 22

 

4x – 2y = 6

+ 3x + 2y = 22

7x + 0 = 28

 

7x = 28

x = 4

 

3(4) + 2y = 22

12 + 2y = 22

2y = 10

y = 5

 

 

3.  2x + 3y = 19

2x + 2y = 14(-1) = - 2x – 2y = - 14

 

- 2x – 2y = - 14

+ 2x + 3y = 19

0 + y = 5

 

y = 5

 

2x + 3(5) = 19

2x + 15 = 19

2x = 4

x = 2

 

4.      4x – 7y = 10(2) = 8x – 14y = 20

5x + 14y = 58

 

8x – 14y = 20

+  5x + 14y = 58

13x + 0 = 78

 

13x = 78

x = 6

 

5(6) + 14y = 58

30 + 14y = 58

14y = 28

y = 2

 

5.     -3x – 2y = - 18(7) = - 21x – 14y = - 126

-4x – 7y = - 37(-2) = 8x + 14y = 74

 

-21x – 14y = -126

+ 8x + 14y = 74

-13x + 0 = -52

 

-13x = -52

x = 4

 

-3(4) – 2y = -18

-12 – 2y = -18

-2y = =6

y = 3

 

 

6.  3x – y = 5n

      4x + y = 9n

 

3x – y = 5n

+ 4x + y = 9n

7x + 0 = 14n

 

7x = 14n

x = 2n

 

4(2n) + y = 9n

8n + y = 9n

y = n

 

7.  2r + p = 23.99

r + 4p = 43.46(-2) = -2r – 8p = - 86.92

 

- 2r – 8p = - 86.92

+ 2r + p = 23.99

0 – 7p = - 62.93

 

-7p = -62.93

p = 8.99

 

r + 4(8.99) = 43.46

r + 35.96 = 43.46

r = 7.50

 

3x3 Elimination:

 

1.     2x + 3y + 4z = 29

4x + 2y – 4z = - 2

5x + 4y – 4z = 6

 

2x + 3y + 4z = 29

+ 4x + 2y – 4z = - 2

6x + 5y + 0 = 27

 

2x + 3y + 4z = 29

+ 5x + 4y – 4z = 6

7x + 7y + 0 = 35

 

6x + 5y = 27(7) = 42x + 35y = 189

7x + 7y = 35(-5) = - 35x – 35y = - 175

 

- 35x – 35y = - 175

+ 42x + 35y = 189

7x + 0 = 14

 

7x = 14

x = 2

 

6(2) + 5y = 27

12 + 5y = 27

5y = 15

y = 3

 

2(2) + 3(3) + 4z = 29

4 + 9 + 4z = 29

4z = 16

z = 4

 

 

2.     x + y + 2z = 15(2) = 2x + 2y + 4z = 30

4x – 2y + z = 22(-2) = - 8y + 4y – 2z = - 44 

3x – 4y + 5z = 15

 

2x + 2y + 4z = 30

+ 4x – 2y + z = 22

6x + 0 + 5z = 52

 

- 8y + 4y – 2z = - 44

+ 3x – 4y + 5z = 15

-5x + 0 + 3z = -29

 

-5x + 3z = -29(6) = - 30x + 18z = - 174

6x + 5z = 52(5) = 30x + 25z = 260

 

- 30x + 18z = - 174

+ 30x + 25z = 260

0 + 43z = 86

 

43z = 86

z = 2

 

6x + 5(2) = 52

6x + 10 = 52

6x = 42

x = 7

 

7 + y + 2(2) = 15

7 + y + 4 = 15

y + 11 = 15

y = 4

 

 

3.    4x + 5y + 8z = 301

3x + 2y – z = 49(8) = 24x + 16y – 8z = 392

5x – 9y + 5z = 12

 

24x + 16y – 8z = 392

+ 4x + 5y + 8z = 301

28x + 21y + 0 = 693

 

3x + 2y – z = 49(5) = 15x + 10y – 5z = 245

 

15x + 10y – 5z = 245

+ 5x – 9y + 5z = 12

20x + y + 0 = 257

 

28x + 21y = 693

20x + y = 257(-21) = - 420x – 21y = -5397

 

- 420x – 21y = -5397

+ 28x + 21y = 693

-392x + 0 = - 4704

 

-392x = - 4704

x = 12

 

28(12) + 21y = 693

336 + 21y = 693

21y = 357

y = 17

 

4(12) + 5(17) + 8z = 301

48 + 85 + 8z = 301

8z = 168

z = 21

 

Quiz:

 

1.     Solve the following with substitution:

 

a.     2x + 3y = 17

x + 2y = 10 = x = - 2y + 10

 

2(-2y + 10) + 3y = 17

-4y + 20 + 3y = 17

-y = -3

y = 3

 

2x + 3(3) = 17

2x + 9 = 17

2x = 8

x = 4

 

b.    3x + 6y = 36 = x = - 2y + 12

4x + 2y = 30

 

4(-2y + 12) + 2y = 30

-8y + 48 + 2y = 30

-6y = -18

y = 3

 

3x + 6(3) = 36

3x + 18 = 36

3x = 18

x = 6

 

c.     2x + 4y = - 6 = x = - 2y - 3

3x – 2y = 23

 

3(-2y – 3) – 2y = 23

-6y – 9 – 2y = 23

-8y = 32

y = - 4

 

2x + 4(-4) = - 6

2x – 16 = -6

2x = 10

x = 5

 

 

d.    1/3x + 3y = 28

2x + 2/3y = 12 = x = - 1/3y + 6

 

1/3(-1/3y + 6) + 3y = 28

- 1/9y + 2 + 3y = 28

2 8/9y = 26

y = 9

 

2x + 2/3(9) = 12

2x + 6 = 12

2x = 6

x = 3

 

2.    Solve the following using elimination:

 

a.  5x + 3y = 25

4x – 3y = -7

 

5x + 3y = 25

+ 4x – 3y = -7

9x + 0 = 18

 

9x = 18

x = 2

 

5(2) + 3y = 25

10 + 3y = 25

3y = 15

y = 5

 

b.     2x + 5y = 46

2x – 4y = -26(-1) = - 2x + 4y = 26

 

- 2x + 4y = 26

+ 2x + 5y = 46

0 + 9y = 72

 

9y = 72

y = 8

 

2x + 5(8) = 46

2x + 40 = 46

2x = 6

x = 3

 

c.     5x – 3y = 26(2) = 10x – 6y = 52

3x + 6y = 0

 

10x – 6y = 52

+ 3x + 6y = 0

13x + 0 = 52

 

13x = 52

x = 4

 

3(4) + 6y = 0

12 + 6y = 0

6y = - 12

y = - 2

 

d.    9x + 8y = 9.5(-5) = - 45 – 40y = - 47.5

5x + 6y = 5.9(9) = 45x + 54y = 53.1

 

- 45 – 40y = - 47.5

+ 45x + 54y = 53.1

0 + 14y = 5.6

 

14y = 5.6

y = 0.4

 

9x + 8(0.4) = 9.5

9x + 3.2 = 9.5

9x = 6.3

x = 0.7

 

 

 

e.    4x + 2y – z = 17

3x – 2y + 4z = 6

x – 2y + 9z = 17

4x + 2y – z = 17

+ 3x – 2y + 4z = 6

7x + 0 + 3z = 23

 

4x + 2y – z = 17

+ x – 2y + 9z = 17

5x + 0 + 8z = 34

 

7x + 3z = 23(-5) = -35x – 15z = - 115

5x + 8z = 34(7) = 35x + 56z = 238

 

-35x – 15z = - 115

+ 35x + 56z = 238

0 + 41z = 123

 

41z = 123

z = 3

 

7x + 3(3) = 23

7x + 9 = 23

7x = 14

x = 2

 

4(2) + 2y – 3 = 17

8 + 2y – 3 = 17

2y = 12

y = 6

 

f.       2x – 4y + 3z = - 29(5) = 10x – 20y + 15z = - 145

4x – 6y + 5z = - 41(-3) = - 12x + 18y – 15z = 123

5x – 8y + 7z = - 55

 

- 12x + 18y – 15z = 123

+ 10x – 20y + 15z = - 145

-2x – 2y + 0 = - 22

 

4x – 6y + 5z = - 41(7) = 28x – 42y + 35z = - 287

5x – 8y + 7z = - 55(-5) = - 25x + 40y – 35z = 275

 

- 25x + 40y – 35z = 275

+ 28x – 42y + 35z = - 287

3x – 2y + 0 = -12

 

-2x – 2y = - 22

3x – 2y = -12(-1) = - 3x + 2y = + 12

 

-2x – 2y = - 22

+ - 3x + 2y = +12

-5x + 0 = - 10

 

-5x = - 10

x = 2

 

3(2) – 2y = - 12

6 – 2y = -12

-2y = - 18

y = 9

 

5(2) – 8(9) + 7z = - 55

10 – 72 + 7z = - 55

7z = 7

z = 1

 

3.  c + 2m = 6.50 = c = -2m + 6.50

2c + m = 7.00

 

2(-2m + 6.50) + m = 7

- 4m + 13 + m = 7

- 3m = -6

m = 2

 

2c + 2  = 7

2c = 5

c = 2.50

 

 

4.    3b +n = 65(-2) = -6b – 2n = - 130

2b + 2n = 70

-6b – 2n = - 130

+ 2b + 2n = 70

-4b + 0 = -60

 

-4b = -60

b = 15

 

2(15) + 2n = 70

30 + 2n = 70

2n = 40

n = 20

 

5.  2s + 3h + c = 24(-2) = -4s – 6h – 2c = - 48

s + 2h + 2c = 24

2s + 3h = 24

 

-4s – 6h – 2c = - 48

+ s + 2h + 2c = 24

- 3s – 4h + 0 = - 24

 

- 3s – 4h = - 24(2) = - 6s – 8h = - 48

2s + 3h = 17(3) = 6s + 9h = 51

 

- 6s – 8h = - 48

+ 6s + 9h = 51

0 + h = 3

 

h = 3

 

2s + 3(3) = 17

2s + 9 = 17

2s = 8

s = 4

 

4 + 2(3) + 2c = 24

4 + 6 + 2c = 24

2c = 14

c = 7

 

 

Web Sites For Additional Information

 

Assuming you need additional info after this amazing tutorial, here are a few pages to look up:

 

www.askdrmath.com – A good page for any math question.  The archives contain tons of info on pretty much any math related question.

 

www.sosmath.com – Another good math page.  With very detailed information.  I highly recommend it.

 

www.coolmath.com – A site designed for easier math things, but a bit of a help.  Recommended for the games mostly, especially Lunar Lander.

 

Bibliography

 

www.sosmath.com

www.askdrmath.com