Question

What is the solution of the system? Use elimination.

Solve by Multiplying BOTH equations.

3x + 2y = 1
4x + 3y = -2

Answer 1

`\left \{ {{3x+2y=1} \atop {4x+3y=-2}} \right.`

Multiply
`3x+2y=1` by 4
`12 x+8y=4`

Multiply
`4x+3y=-2` by 3
`12x+9y=-6`


`\left \{ {{12x+8y=4} \atop {12x+9y=-6}} \right.`


`\left \{ {{12x+9y=-6} \atop {12x+8y=4}} \right. -`


`y = -10`


`\left \{ {{12x+8y=4} \atop {y=-10}} \right.`

for
`12x+8y=4` plug in
`y=-10`


`12x+8(-10) = 4`


`12x - 80 = 4`

Add 80 to both sides:


`12 x = 4 + 80`


`12 x = 84`

Divide both sides by 12 :


`x = (84)/(12)`


`x = 7`


`x = 7 , y = -10`

hope this helps!

Answer 2

( 3x + 2y = 1 ) * 4
( 4x + 3y = -2 ) *3

12x + 8y = 4 --> equation 1
12x + 9y = -6 --> equation 2
-12x - 9y = 6 ---> equation 3 ( changing signs. )
eliminating the x values and adding the like terms

-y = 10
y = -10

Substituting the value of y in equation 1
3x + 2(-10 ) = 1
3x -20 = 1
3x = 21
x = 7

So, X = 7 and Y = -10

Hope this helps... :)