How do you solve #8(y-5)+2y=-10#?

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Answer 1 · 2016-07-02T20:34:16

#y = 3#

Multiply out the bracket and collect like terms.

#8y - 40 + 2y = -10#

#10y = -10 + 40 = 30#

Divide both sides by 10.

#y = 3#

(Just to check we have obtained the correct answer, sub back into the original equation)

#8(3-5) + 2(3) = -16 + 6 = -10#

So our solution works!