Download Substitution and Elimination

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
3x / 3= (9 + 6y) / 3
x = 3 + 2y
-6y goes to the other side, becoming +6y
Both sides are divided by 3 to eliminate the 3
in front of the x
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
2(3 + 2y) + 3y = 27
6 + 4y + 3y = 27
6 + 7y = 27
7y = 27 – 6
7y / 7= 21 / 7
y=3
Substitute (3 + 2y) where x is
Multiply out the bracket to get your new values
Combine the y terms
Move your 6 to the other side of the equation,
making it –6
Subtract 6 from 27
Divide both sides by seven to get the y value
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
3x – 6(3) = 9
3x – 18 = 9
3x = 9 + 18
3x / 3 = 27 / 3
x=9
Substitute 3 for y
Multiply 6 by 3
Move –18 to the other side of the equation,
making it +18
Add 9 and 18
Divide both sides by three to get the value of x
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
x=3+y
Move the –y to the other side, making it +y
Now substitute that value into the first equation in the place of x and
solve for y:
x + y = 11
(3 + y) + y = 11
3 + y + y = 11
3 + 2y = 11
2y = 11 – 3
2y / 2 = 8 / 2
y=4
Substitute (3 + y) in the place of x
Drop your brackets
Combine your y variables
Move your 3 to the other side, becoming –3
Subtract 3 from 11
Divide both sides by 2 to get the value of y
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
x– 4=3
x=7
Substitute 4 for y
Move –4 to the other side, becoming +4
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
2x / 2 = (28 – 6y) / 2
x = 14 – 3y
Move 6y to the other side, becoming –6y
Divide both sides by 2 to get your x value
Our new x value can now be placed in the first equation and we can
solve for y:
4x + 3y = 20
4(14 – 3y) + 3y = 20
56 –12y + 3y = 20
56 – 9y = 20
- 9y = 20 – 56
- 9y / -9= - 36 / -9
y=4
Substitute (14 – 3y) in place of x
Multiply out your brackets with the 4
Combine your y variables
Move 56 to the other side, becoming -56
Subtract 56 from 20
Divide both sides by –9 to get the y value
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
2x + 6(4) = 28
2x + 24 = 28
2x = 28 – 24
2x / 2 = 4 / 2
x=2
Substitute 4 for y
Multiply 6 by 4
Move 24 to the other side, becoming –24
Subtract 24 from 28
Divide both sides by 2 to get the x value
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
4x + 3(1 2/5 – 3x) = 2 8/15
4x + 4 1/5 – 9x = 2 8/15
-5x + 4 1/5 = 2 8/15
-5x = 2 8/15 – 4 1/5
-5x = 2 8/15 – 4 3/15
-5x / -5 = -1 2/3 / -5
Substitute (1 2/5 – 3x) for y
Multiply out the brackets by 3
Combine the y variables
Move 4 1/5 to the other side, becoming
–4 1/5
Find a common denominator for 2 8/15
and 4 1/5
Subtract the two terms
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
6(1/3) + 2y = 2 4/5
2 + 2y = 2 4/5
2y = 2 4/5 – 2
2y / 2 = 4/5 / 2
y = 2/5
Substitute 1/3 for x
Multiply 6 by 1/3
Move 2 to the other side, becoming –2
Combine the two terms
Divide both sides by 2
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
5x / 5 = (7 – 5y) / 5
x = 1.4 –y
Move 5y to the other side, becoming –5y
Divide both sides by 5 to get the x value
Now substitute the new x value into the second equation:
10x – 6y = 4.4
10(1.4 – y) – 6y = 4.4
14 – 10y – 6y = 4.4
14 – 16y = 4.4
-16y = 4.4 – 14
-16y / -16 = -9.6 / -16
y = 0.6
Substitute (1.4 – y) for x
Multiply out the brackets by 10
Combine the y variables
Move 14 to the other side, becoming –14
Combine the two terms
Divide both sides by –16 to get the y value
Now substitute your new value into one of the first two equations. We
will use the first:
5x + 5y = 7
5x + 5(0.6) = 7
5x + 3 = 7
5x = 7 – 3
5x / 5 = 4 / 5
x = 0.8
Substitute 0.6 for y
Multiply 5 by 0.6
Move 3 to the other side, becoming –3
Combine the two terms
Divide both sides by 5 to get the x value
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
x=3
Divide both sides by 6 to get the x value
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
2(3) + 2y = 14
6 + 2y = 14
2y = 14 – 6
2y / 2 = 8 / 2
y=4
Substitute x for 3
Multiply 2 by 3
Move 6 to the other side, becoming –6
Combine the terms
Divide both sides by 2 to get the y value
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)
- 3x – 4y = - 27
Multiply the equation by –1
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
x=5
Divide both sides by 2 to get the x value
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
3(5) + 4y = 27
15 + 4y = 27
4y = 27 – 15
4y / 4 = 12 / 4
y=3
x=5
y=3
Substitute 5 for x
Multiply 3 by 5
Move 15 to the other side, becoming –15
Combine the terms
Divide both sides by 4 to get the y value
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
y=6
Divide both sides by 5 to get the value of y
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
4x + 6 = 34
4x = 34 – 6
4x / 4 = 28 / 4
x=7
Substitute 6 for y
Move 6 to the other side, becoming – 6
Combine the terms
Divide both sides by 7 to get the value of x
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)
-6x – 2y = - 20
Multiply the equation by -2
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
3x + 2(7) = 20
3x + 14 = 20
3x = 20 – 14
3x / 3 = 6 / 3
x=2
Substitute 7 for y
Multiply 2 by 7
Move 14 to the other side, becoming –14
Combine the terms
Divide both sides by 3 to get the x value
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)
63x + 18y = 225
Multiply the equation by 9
(3x + 9y = 27)(-2)
-6x – 18y = - 54
Multiply the equation by -2
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
x=3
Divide both sides by 57 to get the x value
Place the new x value into one of the first two equations and solve for
x. We will use the first:
7x + 2y = 25
7(3) + 2y = 25
21 + 2y = 25
2y = 25 – 21
2y / 2 = 4 / 2
y=2
Substitute x with 3
Multiply 7 by 3
Move 21 to the other side, becoming –21
Combine the terms
Divide both sides by 2 to get the y value
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)
-21x – 6z = - 87
Multiply the equation by –3
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
5(3) + 6z = 39
15 + 6z = 39
6z = 39 – 15
6z / 6 = 24 / 6
z=4
Substitute 3 for x
Multiply 5 by 3
Move 15 to the other side, becoming –15
Combine the terms
Divide both sides by 6 to get the z value
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
2(3) + 3y + 2z = 20
2(3) + 3y + 2(4) = 20
6 + 3y + 8 = 20
3y + 14 = 20
3y = 20 – 14
3y / 3 = 6 / 3
y=2
Substitute 3 for x
Substitute 4 for z
Multiply out both brackets
Combine like terms
Move 14 to the other side, becoming –14
Combine the terms
Divide both sides by 3 to get the y value
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)
-2x – 4y – 6z = - 58
Multiply the equation by –2
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)
4x – 6y + 8z = 50
Multiply the equation by 2
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)
- 7y + 91z = 525
Multiply the equation by 7
And combine with equation four:
- 7y + 91z = 525
+ 7y + 2z = 33
0 + 93z = 558 or 93z = 558
Solve for z:
93z / 93 = 558 / 93
z=6
Divide both sides by 92 to get the z value
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
7y + 2(6) = 33
7y + 12 = 33
7y = 33 – 12
7y / 7 = 21 / 7
y=3
Substitute 6 for z
Multiply 2 by 6
Move 12 to the other side, becoming –12
Combine the terms
Divide both sides by 7 to get the v value
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
x + 2(3) + 3z = 29
x + 2(3) + 3(6) = 29
x + 6 + 18 = 29
x + 24 = 29
x = 29 – 24
x=5
And the answer is:
x=5
y=3
z=6
Substitute 3 for y
Substitute 6 for z
Multiply out the brackets
Combine like terms
Move 24 to the other side, becoming –24
Combine the terms
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)
60x – 25y + 30z = 65
Multiply the equation by 5
(14x + 8y – 5z = 5)(6)
84x + 48y – 30z = 30
Multiply the equation by 6
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)
48x – 20y + 24z = 52
Multiply the equation by 4
(18x – 3y – 4z = - 2)(6)
108x – 18y – 24z = - 12
Multiply the equation by 6
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)
5472x + 874y = 3610
Multiply the equation by 38
(156x – 38y = 40)(23)
3588x - 874y = 920
Multiply the equation by 23
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
144(0.5) + 23y = 95
72 + 23y = 95
23y = 95 – 72
23y / 23 = 23 / 23
y=1
Substitute 0.5 for x
Multiply 144 by 0.5
Move 72 to the other side, becoming –72
Combine the terms
Divide both sides by 23 to get the y value
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
12(0.5) – 5y + 6z = 13
12(0.5) – 5(1) + 6z = 13
6 – 5 + 6z = 13
Substitute 0.5 for x
Substitute 1 for y
Multiply out the brackets
Combine like terms
6z + 1 = 13
6z = 13 – 1
6z / 6 = 12 / 6
z=2
Move 1 to the other side, becoming –1
Combine the terms
Divide both sides by 6 to get the z value
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