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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

User Drneel
by
6.2k points

2 Answers

4 votes

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

User Samuel Heaney
by
6.4k points
4 votes

Answer:

x = 5 , y = 2

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.

User Serikov
by
6.5k points
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