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SYSTEMS OF EQUATIONS
Method #2- Elimination
Previously, we learned how to solve systems of equations by
using substitution. Why did we learn that method first?
Because it’s the most difficult! Let’s learn the easiest…elimination.
Example 1:
2x  3 y  12
This system of equations could be solved
by eliminating the y variable through
x  3 y  6 addition.
_____________
3x
= 18
x=6
2(6) + 3y = 12
12 + 3y = 12
3y = 0
(6,0)
y=0
Then, you would
plug the x in just
like you did with
substitution.
Example 2:
3x + 2y = 16
+
3x – y = 10
_____________
3y = 6
y=2
3x + 2(2) = 16
3x + 4 = 16
3x = 12
x=4
(4, 2)
You can solve this one by multiplying one
of the equations by a negative one.
Again, to finish this problem, you would
plug the y back into one of the
equations.
Why did we put the 4 first in the
parentheses?
Always write your answers in alphabetical
order!
Example 3: Here’s one for you to practice on:
- 3x –+ 2y
= -14
5x
– 2y = 22
_____________
2x
=8
x=4
The solution is (4, -1).
3x – 2y = 14
3(4) – 2y = 14
12 – 2y = 14
-2y = 2
y = -1
Example 4: Here’s another:
3x + 5y = 10
-3x
– 5y = 2
_____________
0 = 12
There are no solutions.
Solve the following systems of equations by using elimination.
1) 2x + 4y = 18
x – 4y = -15
2) 3x – y = -3
3x + 4y = -18
3) 6x + 3y = -3
2x + 3y = -5
Answers:
1) (1, 4)
2) (-2, -3)
3) (1/2, -2)
Example 5: Since this system of equations can’t be solved by
elimination with addition, how can we solve it?
-3(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
added to the top equation
eliminating the variable x.
-3x – 12y = -51
3x – 2y =
9
-14y = -42
y=3
x + 4(3) = 17
x + 12 = 17
x=5
(5, 3)
x  4 y  17
3x  2 y  9
We solved this system of
equations by multiplying the top
equation by 3.
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.
6x  4 y  18
4x  3 y  12
4x – 3y = 12
-4( x  2 y  14)
(+) -4x – 8y = - 56
 11y  44
Multiply the bottom equation by -4 to
get a new system of equations.
Add the bottom equation to the top
equation.
Solve for y.
Solve for x by substituting
the value for y into one of
the equations.
11 y 44

11 11
y4
x  2 y  14
x  2( 4 )  14
x  8  14
x  6 (6, 4)
Example 7: Since this system of equations can’t be solved by
multiplying one row by a number, how can we solve it?
-3( 2x + 3y = 6 )
2( 3x – 4y = -25)
If the top equation was
multiplied by -3, then the first
term would be -6x. Then
multiply the bottom equation
by 2, and the first term
becomes a 6x.
-6x – 9y = -18
6x – 8y = -50
-17y = -68
y=4
2x + 3(4) = 6
2x + 12 = 6
2x = -6
x = -3
(-3, 4)
Example 8: Since this system of equations can’t be solved by
multiplying one row by a number, how can we solve it?
-3( 7x + 2y = 4 )
2( 5x + 3y = 6 )
-21x – 6y = -12
10x + 6y = 12
-11x = 0
x=0
7(0) + 2y = 4
0 + 2y = 4
2y = 4
y=2
(0, 2)
Solve the following systems of equations by using elimination.
1) 2x + y = 6
x – 4y = -15
2) 3x – y = -3
2x + 5y = -19
4) 3x + 5y = 7
6x + y = 5
6) 3x – 2y = -10
5x + 2y = 10
Answers:
3) x + 2y = -1
2x + 3y = 2
5) 2x + 3y = 1 Page 269/ 7-15 odd
3x + 4y = 2
7) 3x + 4y = 0
8) 5x – 3y = -27
5x – 4y = -8
1) (1, 4) 2) (-2, -3)
5) (2, -1) 6) (0, 5)
Page 276/ 7-15 odd
3) (7, -4)
7) (-1, ¾)
2x + 7y = 22
4) (2/3, 1)
8) (-3, 4)