how to solve 25-[2+5y-3(y+2)]=-3(2y-5)-[5(y-1)-3y+3]

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1 Answer

25 - (2 + 5y - 3(y+2) ) = -3 (2y - 5) - (5(y-1) - 3y + 3)

25 - 2 - 5y + 3(y+2) = -3 (2y - 5) - (5(y-1) - 3y + 3)

25 - 2 - 5y + 3y + 6 = -3 (2y - 5) - (5(y-1) - 3y + 3)

23 - 2y + 6 = -3 (2y - 5) - (5(y-1) - 3y + 3)

-2y + 29 = -3 (2y - 5) - (5(y-1) - 3y + 3)

-2y + 29 = -6y + 15 - (5(y-1) - 3y + 3)

-2y + 29 = -6y + 15 -5(y-1) + 3y - 3

-2y + 29 = -6y + 15 -5y + 5 + 3y - 3

-2y + 29 = -8y + 15 + 5 - 3

-2y + 29 = -8y + 17

6y + 29 = 17

6y = -12

y = -2
by Level 13 User (103k points)

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