Objective
The student will be able to:
solve systems of equations using
elimination with multiplication.
Solving Systems of Equations
So far, we have solved systems using
graphing, substitution, and elimination.
This goes one step further and show how
to use ELIMINATION with multiplication.
What happens when the coefficients are
not the same?
We multiply the equations to make them
the same! You’ll see…
1) Solve the system using elimination.
2x + 2y = 6
3x
–
y
=
5
Multiply the bottom
equation by 2
2x + 2y = 6
(2)(3x – y = 5)
2x + 2y = 6
(+) 6x – 2y = 10
8x
= 16
x=2
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y=1
(2, 1)
2) Solve the system using elimination.
x + 4y = 7
4x – 3y = 9
Multiply the top equation by -4
(-4)(x + 4y = 7)
-4x – 16y = -28
(+) 4x – 3y = 9
4x – 3y = 9
-19y = -19
4x – 3y = 9
y=1
4x – 3(1) = 9
4x – 3 = 9
4x = 12
x=3
(3, 1)
Which variable is easier to eliminate?
3x + y = 4
4x + 4y = 6
 y is easier to eliminate.
 Multiply the 1st equation by (-4)
3) Solve the system using elimination.
3x + 4y = -1
4x – 3y = 7
Multiply both equations
(3)(3x + 4y = -1)
9x + 12y = -3
(+) 16x – 12y = 28
(4)(4x – 3y = 7)
25x = 25
x=1
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
(1, -1)
4) What is the best number to multiply the top
equation by to eliminate the x’s?
3x + y = 4
6x + 4y = 6
 Multiply the 1st equation by (-2)
 (-2)(3x + y = 4)  -6x -2y = -8
6x + 4y = 6
2y = -2
y=1
(1/3, 1)
6x + 4y = 6
6x + 4(1) = 6
6x = 2
x = 2/6
x = 1/3
5) Solve using elimination.
2x – 3y = 1
x + 2y = -3
 Multiply 2nd equation by (-2) to
eliminate the x’s.
 (-2)(x + 2y = -3)  -2x – 4y = 6
1st equation 
2x – 3y = 1
-7y = 7
y = -1
2x – 3y = 1
2x – 3(-1) = 1
2x + 3 = 1
2x = -2
x = -1
(-1, -1)
5) Solve using elimination.
2(4x + y = 6 )
-8x – 2y = 13
8x + 2y = 12
-8x – 2y = 13
0 = 25
Add down to eliminate x.
But look what happens, y is
eliminated too.
We now have a false statement,
thus the system has no
solution, it is inconsistent.
6) Elimination
Solve the system using elimination.
5x – 2y = –15(4) Since neither variable will drop out if the
equations are added together.
3x + 8y = 37
20x – 8y = –60
3x + 8y = 37
23x = –23
x = –1
Multiply one or both of
the equations by a constant to make one of
the variables have the same number with
opposite signs.
The best choice is to multiply the top
equation by 4 since only one equation would
have to be multiplied.
Also, the signs on the y-terms are
already opposites.
6) Elimination
x = –1
3x + 8y = 37 (second equation)
3(–1) + 8y = 37
–3 + 8y = 37
8y = 40
y=5
The solution is (–1, 5)
To find the second
variable, substitute in any
equation that contains two
variables.
7) Elimination
Solve the system using elimination.
4x + 3y = 8 (5) For this system, we must multiply both
3x – 5y = –23 (3) equations by a different constant in order to
make one of the variables “drop out.”
20x + 15y = 40
9x – 15y= –69
29x = –29
x = –1
It would work to multiply the top equation
by –3 and the bottom equation by 4 OR to
multiply the top equation by 5 and the
bottom equation by 3.
7) Elimination
x = –1
4x + 3y = 8
4(–1) + 3y = 8
–4 + 3y = 8
3y = 12
y=4
The solution is (–1, 4)
8) Elimination Method
Solve the system.
3x  4 y  1
2 x  3 y  12
(1)
(2)
To eliminate x, multiply equation (1) by –2 and
equation (2) by 3 and add the resulting equations.
 6 x  8 y  2
6 x  9 y  36
17 y  34
y2
8) Elimination Method
Substitute 2 for y in (1) or (2).
3 x  4(2)  1
3x  9
x3
The solution is (3, 2)
9) Solving an Inconsistent System
Solve the system
3x  2 y  4
 6x  4 y  7
(1)
(2)
Eliminate x by multiplying (1) by 2 and
adding the result to (2).
6x  4 y  8
 6x  4 y  7
0  15
Solution set is .
Inconsistent System
10) Solving a System with Dependent
Equations
Solve the system.
 4x  y  2
8x  2 y   4
(1)
(2)
Eliminate x by multiplying (1) by 2 and adding the
result to (2).  8 x  2 y  4
8 x  2 y  4
00
Each equation is a solution of the other.
Infinite solutions.