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Solving Systems of Equations by ELIMINATION
In Chapter 2 we learned to solve systems of equations by COMPARISON. It is possible to
solve those same systems and get the same answers by ELIMINATION.
Sometimes it is an easier method to use.
Linear equations having 2 variables can be presented in 2 common forms:
Slope y-intercept Form
y = mx + b
Standard Form
Ax + By = C
To express an equation in the Slope y-intercept form, we must isolate y. Slope yintercept form is used because it is easy to determine the slope and y-intercept of a line
on a graph from this form.
When both equations from a system are presented this way, the COMPARISON method
is easier to use.
To express an equation in Standard form, we must place the variables on the left side of
the equation and the numeric terms on the right. This does not require the use of
fractions.
When both equations are presented this way, the ELIMINATION method is easier to use.
Whichever method is used, the 2 equations containing 2 variables are combined in such
a way as to get 1 equation that contains only 1 variable. This means that either the x
variable or the y variable must be eliminated.
We can get the idea of how elimination works by starting with some simple examples.
-6x + 5y = 1 (1)
6x + 4y = -10 (2)
When we say elimination, we mean that we eliminate one of the variables by adding the
equations together.
Notice that the coefficient for x in equation (1) is -6 and
-6x + 5y = 1 (1)
the coefficient for x in equation (2) is 6. They are
6x + 4y = -10 (2)
opposite and that means when we add them they will
cancel each other out.
9y = -9
9y
9
Now we have an equation that has only one variable –

9
9
the y variable in this case. We can find the value of y.
y = -1
After finding y, we can use its value with one of the
original equations to find x.
-6x + 5y = 1 (1)
6x + 4y = -10 (2)
6x + 4(-1) = -10
6x – 4 = -10
6x = -10 + 4
6x = -6
-6x + 5(-1) = 1
-6x – 5 = 1
-6x = 1 + 5
-6x = 6
 6x
6

6
6
 6x
6

6
6
x = -1
(-1,-1)
x = -1
It doesn’t matter which equation we
choose, we will find the same result for
x. Using the second equation is a way
to verify our answer.
In the first example, the x variable was eliminated just by adding the equations.
Sometimes when equations are added neither variable will be eliminated.
However it is possible to adjust either one or both equations so that when they are
added, one of the variables will be eliminated.
-4x + y = -14 (1)
-4x + y = -14 (1)
5x + 2y = 11 (2)
x + 3y = -3
5x + 2y = 11 (2)
Here when we add the 2 equations neither the x nor
the y variable cancels. This is of no help to us if we
wish to solve the system.
The necessary condition to cancel a variable is that the coefficients of that variable must
be opposite. That is not the case in the situation above.
coefficient of x variable in equation (1)
coefficient of x variable in equation (2)
-4
5
coefficient of y variable in equation (1)
coefficient of y variable in equation (2)
1
2
After analyzing both pairs of coefficients, we can see that it would be relatively easy to
change the coefficient of the y variable in equation (1) so that it is -2. This can be done by
multiplying equation (1) by -2.
-4x + y = -14 (1)
5x + 2y = 11 (2)
x + 3y = -3
Here when we add the 2 equations neither the x nor
the y variable cancels. This is of no help to us if we
wish to solve the system.
The necessary condition to cancel a variable is that the coefficients of that variable must
be opposite. That is not the case in the situation above.
coefficient of x variable in equation (1)
coefficient of x variable in equation (2)
-4
5
coefficient of y variable in equation (1)
coefficient of y variable in equation (2)
1
2
After analyzing both pairs of coefficients, we can see that it would be relatively easy to
change the coefficient of the y variable in equation (1) so that it is -2. This can be done by
multiplying equation (1) by -2.
( -4x + y = -14 (1) ) × -2
8x – 2y = 28 (1)
Notice that now the coefficient of y is opposite for in both equations.
8x – 2y = 28 (1)
5x + 2y = 11 (2)
13x = 39
13
13
x =3
Now when we add them the y-variable cancels
and we can solve for x.
Notice that now the coefficient of y is opposite for in both equations.
8x – 2y = 28 (1)
5x + 2y = 11 (2)
Now when we add them the y-variable cancels
and we can solve for x.
5x + 2y = 11 (2)
-4x + y = -14 (1)
13x = 39
5(3) + 2y = 11 (2)
13
13
-4(3) + y = -14 (1)
15 + 2y = 11
x =3
-12 + y = -14
2y = 11 – 15
y = -14 + 12
2y = -4
y = -2
y = -2
(3,-2)
Sometimes the system requires that both equations be adjusted so as to cause one of
the variables to cancel.
4x + 3y = -9 (1)
14x + 4y = -12 (2)
coefficient of x variable in equation (1) 4
coefficient of x variable in equation (2) 14
coefficient of y variable in equation (1) 3
coefficient of y variable in equation (2) 4
If we add the x coefficients in these 2 equations, we get 18x and if we add the ycoefficients, we get 7y. Also there is no way to multiply one equation by an integer to
make either pair of coefficients opposite.
In situations like this, we must determine the lowest common multiple (LCM) of one of
the pairs of coefficients. Let’s use the x coefficients: 4 and 14. The LCM for these
28 ÷ 4 = 7
numbers is 28 because both numbers can divide into 28.
28 ÷ 14 = 2
4x + 3y = -9 (1)
14x + 4y = -12 (2)
Due to the fact that the LCM for 4 and 14 is 28, we need to multiply one equation by a
number to make a coefficient of +28 and the other by a number to make -28.
(
4x + 3y = -9 (1) )
28x + 21y = -63 (1)
28x + 21y = -63 (1)
-28x – 8y = 24 (2)
13y = -39
13
13
y = -3
×7
( 14x + 4y = -12
(2))
× -2
-28x – 8y = 24 (2)
4x + 3y = -9 (1)
4x + 3(-3) = -9 (1)
4x – 9 = -9
4x = -9 + 9
4x = 0
x=0
(0,-3)
We can just as easily adjust the equations so as to cancel the y-coefficients and this will
produce the same solution to the system. 4x + 3y = -9 (1)
14x + 4y = -12 (2)
coefficient of y variable in equation (1)
coefficient of y variable in equation (2)
3
4
In this situation, the lowest common multiple (LCM) of the y coefficients: 3 and 4 is 12.
They can both divide into 12.
12 ÷ 3 = 4
12 ÷ 4 = 3
Due to the fact that the LCM for 3 and 4 is 12, we need to multiply one equation by a
number to make a coefficient of +12 and the other by a number to make -12.
(
4x + 3y = -9 (1) )
16x + 12y = -36 (1)
16x + 12y = -36 (1)
-42x – 12y = 36 (2)
-26x = 0
-26 -26
x =0
×4
( 14x + 4y = -12
(2))
× -3
-42x – 12y = 36 (2)
4x + 3y = -9 (1)
4(0) + 3y = -9 (1)
0 + 3y = -9
3y = -9
y = -3
(0,-3)
Finally, when we try to cancel one variable, the other variable may cancel as well as
shown in the next example.
-2x + 3y = 4 (1)
6x – 9y = 1 (2)
In this situation, we can multiply equation (1) by 3 to make the coefficient of the x
variable in equation (1) opposite to that in equation (2).
(
-2x + 3y = 4 (1) )
-6x + 9y = 12 (1)
6x – 9y = 1 (2)
×3
When the equations are added not only do the
x variables cancel but so do the y variables.
0 = 13
The result of the addition of the 2 equations is a statement. in this case – a FALSE
statement. This indicates that there is no solution to this system. If a graph were
constructed of this system, the lines for each equation would be parallel and never
intersect.
If the result of the addition of the 2 equations had been a TRUE statement, it would
indicate that there are infinite solutions to this system. If a graph were constructed, the
lines for each equation would be coinciding and every point on the line would be
considered part of the solution.
Summary of Solving Systems of Equations by ELIMINATION
y = 2x + 3 (1)
3x + 4 = 2y (2)
Step 1: Transform both equations to standard form.
-2x + y = 3 (1)
3x + -2y = -4 (2)
Step 2: Choose which variable you wish to cancel.
It would be easier to cancel the y-variable in this particular case
Step 3: Adjust one or both equations so as the coefficients of the variables you wish to
cancel are opposite.
( -2x + y = 3 (1) ) × 2
-4x + 2y = 6 (1)
-4x + 2y = 6 (1)
3x + -2y = -4 (2)
Step 4: Add both equations together.
-x = 2
x = -2
Step 5: Use the value of the variable that was determined in Step 4 to calculate the value
of the other variable with one of the original equations.
y = 2x + 3 (1)
y = 2(-2) + 3 (1)
y = -4 + 3
y = -1
Step 6: Put the solution in the form of an ordered pair.
(-2,-1)