Question

(06.01) 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

we have that

x + 3y = 6 -------> first equation

2x − y = 10

x + 3y = 6-----> substract x both sides

-x+x+3y=6-x

3y=6-x-----> divide by 3 both sides

y=(6-x) /3

y=2-(x/3)

y=-(x/3)+2

therefore

the answer is

negative one thirdx + 2