How do you solve #-12= 3( y + 5)#?

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Answer 1 · 2018-04-20T00:12:51

y = -9

First distribute the "3" to everything inside of the parentheses
-12 = 3y + 15

Then combine like terms by subtracting both sides by 15
-12 - 15 = 3y + 15 - 15
-27 = 3y

Now divide both sides by 3, in order to isolate the variable
#-27/3# = #3y/3#

-9 = y

Hope this made sense

Answer 2 · 2018-04-20T00:13:55

#y=-9#

Solve:

#-12=3(y+5)#

Expand the right-hand side using the distributive property: #a(b+c)=ab+ac#.

#-12=3y+15#

Subtract #15# from both sides.

#-15-12=3y color(red)cancel(color(black)(+15))color(red)cancel(color(black)(-15))#

Simplify.

#-27=3y#

Divide both sides by #3#.

#(-27)/3=(color(red)cancel(color(black)(3))y)/color(red)cancel(color(black)(3))#

Simplify.

#-9=y#

Switch sides.

#y=-9#