What is the y-value in the solution to this system of linear equations? 4x + 5y = -12
-2x + 3y = -16
-4.
-2

2
5

y = -4

Step-by-step explanation:

4x + 5y = -12 eq1

-2x + 3y = -16 ...eq2

From eq1, solve for x:

4x + 5y = -12

4x = -12 - 5y

x = -12 - 5y/4

From eq2, substitute value of x:

-2(-12-5y/4) + 3y = -16

3y - 2 (-5y-12/4) = -16

3y - 2(-5y-12)/4 = -16

12y - 2(-5y - 12) = -16

4*3y - 4*2(-5y-12)/4 = 4*(-16)

12y - 2(-5y-12) = -64

12y + 10y + 24 = -64 (divide both sides by common factor 2)

6y + 5y + 12 = -32

11y = -32 - 12

11y = -44

Divide both sides by 11

11y/11 = -44/11

y = -4

12/11c

or

12

c

11

the answer is simply just 6n add the two together as they have an n