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

just put them in order

step-by-step explanation: