What is the solution to this system of equations? 5x + 2y = 29 x + 4y = 13 A. x = 5, y = 3 B. x = 2, y = 5 C. x = 5, y = 2 D. x = 3, y = 2

5x + 2y = 29

x + 4y = 13

First isolate one of the variables in one of the equations.

I will isolate x in the second equation (because it is the easiest to isolate) and plug it into the second equation

x + 4y = 13 Subtract 4y on both sides

x = 13 - 4y

5x + 2y = 29

5 (13 - 4y) + 2y = 29 [since x = 13 - 4y, you can plug in (13 - 4y) for x]

Multiply/distribute 5 into (13 - 4y)

65 - 20y + 2y = 29

65 - 18y = 29 Subtract 65 on both sides

-18y = - 36 Divide - 18 on both sides

y = 2

Now that you know y, you can plug it into one of the equations to find x

x + 4y = 13

x + 4 (2) = 13

x + 8 = 13

x = 5

Your answer is C

x = 5, y = 2

Step-by-step explanation:

5x + 2y = 29 ... (1)

x + 4y = 13 Multiply this equation by - 5:-

-5x - 20y = - 65 ... (2)

Add equations (1) and (2) : -

-18y = - 36

y = 2

Substitute for y in equation (1) : -

5x + 2 (2) = 29

5x = 25

x = 5

Check the results in the second equation:-

x + 4y = 13

5 + 4 (2) = 13 - It checks out.