Question

I am in like desperate need of help...

In the below system, solve for y in the first equation.

x + 3y = 6
2x − y = 10

A. one thirdx + 2

B. negative one thirdx + 6

C. −x + 2

D. negative one thirdx + 2

Answer 1

@mathmale @whpalmer4

Answer 2

\[x+3y=6\]to solve for \(y\) we need to get it alone on one side of the equation. We can do just about anything to accomplish this, so long as we do it to each side of the equation. Let's start by subtracting \(x\):
\[x+3y-x=6-x\]\[3y=6-x\]

Answer 3

At this point, you should divide both sides of the equation by 3, which will give us \[\frac{3y}{3}=\frac{\cancel{3}y}{\cancel{3}}=y\]for the left side of our equation. Whatever would be on the right hand side after we divided both sides by 3 is the answer you seek.

Answer 4

Thank you for your help! I am so sorry I did not respond. I actually ended up figuring out how to do this problem after I asked for help. But thank you again! <3