How do you solve #\frac { 1} { 6} ( y + 42) - 15= - 3#?

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Answer 1 · 2016-12-20T12:49:34

#y = 30#

First, expand the terms within parenthesis and combine like terms:

#1/6y + 42/6 - 15 = -3#

#1/6y + 7 - 15 = -3#

#1/6y - 8 = -3#

Next, isolate the #y# term on one side of the equation, the constants on the other side of the equation while keeping the equation balanced:

#16y - 8 + color(red)(8) = -3 + color(red)(8)#

#1/6y - 0 = 5#

#1/6y = 5#

Now, solve for #y# while keeping the equation balanced:

#color(red)(6) xx 1/6y = color(red)(6) xx 5#

#6/6y = 30#

#1y = 30#

#y = 30#