How do you solve #2y+18=12-6(y+7)#?

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Answer 1 · 2015-10-28T11:52:37

#y=-8#

Expanding the brackets:
#2y+18=12-6y-42#

Adding the constants on the right and removing the constant from the left.

Subtract 18 from both sides
#(2y+18) - 18 =(-6y-30)-18#

#2y = -6y - 48#

Moving the 'y' term from the right

Add #6y# to both sides

#(2y)+6y=(-6y-48) + 6y#

#8y=-48#

To remove the 8 from #8y#

Divide both sides by 8 (divide by 8 is the same multiply by #1/8#)
#(8y)/8 =(-48)/8#

#8/8 y = - 8#

#y=-8#