How do you solve #8y + 30= 2( 4y + 3)#?

1 Answer
May 12, 2018

No Solution or #cancel0#.

Explanation:

First, use the distributive to simplify #color(blue)(2(4y+3))#:
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Following this image:
#color(blue)(2(4y+3) = (2 * 4y) + (2 * 3) = 8y + 6)#

Let's put that back into the equation:
#8y + 30 = 8y + 6#

Now subtract #color(blue)(8y)# from both sides:
#8y + 30 quadcolor(blue)(-quad8y) = 8y + 6 quadcolor(blue)(-quad8y)#

#30 = 6#

Oh no! Our variable is gone! This means we have to look at our equation and see if it is true.

Is it true that #30 = 6#? No. That means there is No Solution, or #cancel0#.