Now that we have one of the variables from Part A isolated, and written in terms of the other variable, we can now substitute this into the other of the two original equations.

2A + 3(3A-5) = -4

We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one of the original equations (or the expression derived in Part A) to solve for the other variable. When this is done with the system of two equations from Parts A and B, what is the solution?

Complete Question

The complete question is shown on the first second and third uploaded image

The first uploaded image is part A

The second uploaded image is part B

The third uploaded image is part C

A = 1  and B = -2

Step-by-step explanation:

The explanation is shown on the fourth uploaded image

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1