Question

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

x + 3y = 6
2x – y = 10

one thirdx + 2

negative one thirdx + 6

–x + 2

negative one thirdx + 2

Answer 1

make x+3y=6 into x=6-3y
use substitution, plug in 6-3y into the 2nd equation anywhere there is an x
After you solve that 2nd equation for y, plug in the answer for y into the place in the 1st equation where there is a y and you will then solve for x.

Answer 2

would the answer be negative one thirdx + 2 ? @IndianChick500