Download Elimination Using Addition and Subtraction

Previously, we learned that adding two numbers together which
have the same absolute value but are opposite in sign results in
a value of zero. This can be shown graphically using algebra
tiles.
Each algebra tile has the same
1
1
1
absolute value but is opposite in
sign. Each pair of tiles makes a
-1
-1
-1
zero pair.
This same principle can be applied to variables. Below is a
zero pair for the variable, “x”.
x
-x
We can apply the idea of a zero pair to systems of equations so
that one of the variables can be eliminated.
Look at the following system of equations. What do you
observe?
x y6
x y2
Zero pair
In each equation, there is a “y”. However, the signs in
the two equations are opposite. If added together, the “y”
and the “-y” would make a zero pair.
Add the following two
equations together by
adding like terms
together.
Now pick one of the two
equations and substitute the
value of “x” into that
equation and solve for “y”.
x y2
2x  0  8
x+y=6
x y6
Now solve for “x”.
2x = 8
2
2
x=4
4+y=6
-4
-4
y=2
Write the solution as an
ordered pair.
Solution:
(4,2)
x y6
x y2
Solution:
(4,2)
To check your work, substitute the values for the variables into
each equation and determine if it is true.
x y6
x y2
4 2 6
66
4 2 2
2 2

The solution is correct.

Addition worked when the signs of one of the variables were
opposite. However, you may encounter a system of equations in
which the signs of the variables are the same. In this case,
instead of adding, you will be subtracting one equation from the
other. Remember that subtraction is really the same as adding
the opposite.
Let’s take a look at the following system of equations:
4x  y  9
3x  y  6
The signs of the variables are all positive. Therefore, in
order to solve this system, we can subtract the bottom
equation from the top equation.
Subtract the bottom
equation from the top
equation.
4x  y  9
(-)
3x  y  6
x03
or
x3
Since we have the value for
“x”. Pick one of the equations
and substitute the value 3 for
“x” and solve for “y”.
3x + y = 6
(3)(3) + y = 6
9+y=6
-9
-9
y = -3
Write the solution as an
ordered pair.
Solution:
(3,-3)
4x  y  9
3x  y  6
Solution:
(3,-3)
To check your work, substitute the values for the variables into
each equation and determine if it is true.
4x  y  9
4 ( 3)  ( 3)  9
12  3  9
9 9
The solution is correct.
3x  y  6
3( 3)  ( 3)  6
9 3 6
66 
Solve the following system of
equations by using addition or
subtraction.
2x  y  2
2x  y  5
This system of equations can be
solved by using subtraction.
2x  y  2
(-)
2x  y  5
0  0  7
When subtraction is used,
there are no more variables
(x or y) remaining and the
result is an incorrect
statement. We know that
0 7
Therefore, this system
of equations represents
parallel lines and there
is no solution.
1. Arrange the two equations so that the like terms are in vertical
columns.
2. If the signs of the variables are opposite, add the two equations
together to eliminate one of the variables.
3. If the signs of the variables are the same, then subtract one of
the equations from the other equation.
4. Solve for the remaining variable.
5. Substitute the value of the variable into one of the equations
and solve for the other variable.
6. Check the solution by substituting the values of the two
variables into each equation.
7. Write the solution. If there is one solution, write it as an
ordered pair.
Solve each system of equations by using addition or subtraction.
1.
2x  4 y  18
x  4y  3
2.
3 p  2r  5
3 p  6r  15
3. The sum of two numbers is 85. The difference between
the two numbers is 19. What are the two numbers?
2x  4 y  18
x  4y  3
Add the two equations
together.
2x  4 y  18
(+)
x  4y  3
3x  0  21
Solve for “x”.
3x = 21
3 3
x=7
Solve for “y” by substituting
the value for “x” into one of the
equations.
x  4y  3
7  4y  3
7
7
 4 y  4
y1
Solution: (7,1)
3 p  2r  5
3 p  6r  15
Subtract the bottom
equation from the top
equation.
3 p  2r  5
(-)
3 p  6r  15
 4 r  20
Solve for “s” by substituting the
value for “r” into one of the
equations.
3 p  2r  5
3 p  2(5)  5
3 p  10  5
 10  10
3 p  15
Solve for “r”.
4 r 20

4 4
3 p 15

3
3
r  5
p5
Solution:
(5,-5)
The sum of two numbers is 85. The difference between the
two numbers is 19. What are the two numbers?
x  y  85
x  y  19
2x  0  104
2x 104

2
2
x  52
Write an equation for the sum of the numbers.
Write an equation for the difference of the
numbers.
Add the equations together.
Solve for x.
x  y  85
x  y  19
x  52
Solve for “y” by substituting in the
value of “x”.
52  y  85
( 52  52)  y  85  52
y  33
The two numbers are 52 and 33.