(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
1 answer:
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
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