Download Elimination Using Multiplication

x  4 y  17
3x  2 y  9
Previously, we learned how to solve systems of equations by
using addition or subtraction which eliminated one of the
variables.
2x  3 y  12
x  3y  6
3x  2 y  15
3x  y  9
This system of equations could be solved
by eliminating the y variable through
addition.
This system of equations could be
solved by eliminating the x variable
through subtraction.
Since this system of equations can’t be solved by elimination
with addition or elimination with subtraction, how can we solve
it?
x  4 y  17
3x  2 y  9
If the top equation was
multiplied by 3, then the first
term would be 3x. The bottom
equation could then be
subtracted from the top
equation eliminating the
variable x.
3( x  4 y )  3(17)
3x  12 y  51
The new system of
equations is now:
3x  12 y  51
3x  2 y  9
3x  12 y  51
(-)
Subtract the bottom
equation from the top
equation.
3x  2 y  9
14 y  42
14 y 42

14 14
Solve for y.
y3
Solve for x by
substituting the value
for y into one of the
equations.
3x  2 y  9
3x  2( 3)  9
3x  6  9
+ 6 +6
3x  15
3x 15

3
3
x5
x  4 y  17
x5
y3
3x  2 y  9
Substitute the value of the variables into each equation to
determine if the solution is correct.
x  4 y  17
3x  2 y  9
3( 5 )  2( 3)  9
15  6  9
5  4 ( 3)  17
5  12  17
17  17

99
This system of equations represents two lines which intersect at
the point (5,3).

x  4 y  17
3x  2 y  9
Previously, we solved this
system of equations by
multiplying the top equation by
3 and then used elimination by
subtraction.
Could we have used a different factor for the
multiplication?
We could have multiplied the bottom equation by 2 to get
6x  4 y  18
The system of equations would then become
x  4 y  17 Elimination by addition would then be used
to solve this system of equations. The result
6x  4 y  18 should be the same. Try it and see.
1. Arrange the equations with like terms in columns.
2. Multiply one or both equations by an appropriate
factor so that the new coefficients of x or y have the
same absolute value.
3. Add or subtract the equations and solve for the
remaining variable.
4. Substitute the value for that variable into one of the
equations and solve for the value of the other variable.
5. Check the solution in each of the original equations.
Solve the following systems of equations by using elimination.
1.
4x  3 y  12
x  2 y  14
2.
16x  8 y  12
12x  6 y  9
3.
 2x  4 y  3
3x  6 y  8
4x  3 y  12
x  2 y  14
Multiply the bottom equation by 4 to get
a new system of equations.
Subtract the bottom equation from the
top equation.
Solve for y.
4x  3 y  12
(-)
4x  8 y  56
 11y  44
11 y 44

11 11
y4
Solve for x by substituting
the value for y into one of
the equations.
x  2 y  14
x  2( 4 )  14
x  8  14
x6
4x  3 y  12
x6
y4
x  2 y  14
Check the solution by substituting the values for the
variables into each equation.
4 x  3 y  12
4 6  3 4  12
24  12  12
12  12 
x  2 y  14
6  2 4  14
6  8  14
14  14 
This system of equations represents two lines which intersect at the
point (6,4).
16x  8 y  12
12x  6 y  9
The lowest common multiple of 6 and 8
is 24. Multiply the top equation by 3 and
the bottom equation by 4.
3(16x  8 y )  3(12)
48x  24 y  36
4(12x  6 y )  4( 9)
48x  24 y  36
The new system of equations is:
48x  24 y  36
48x  24 y  36
Since these two equations are identical, there is only one line and an
infinite number of solutions.
2x  4 y  3
3x  6 y  8
The lowest common multiple of 2
and 3 is 6. Multiply the top
equation by 3 and the bottom
equation by 2.
3( 2x  4 y )  3( 3)
6x  12 y  9
2( 3x  6 y )  2( 8)
6x  12 y  16
Add the new system of
equations together.
6x  12 y  9
6x  12 y  16
0x  0 y  25
Since 0  25, there is no solution to this system of equations
which represents two parallel lines.