How do you solve #2( 4x + 1) = 3x + 17#?

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Answer 1 · 2016-11-23T17:33:48

#x=3#.

First distribute through the parentheses on the left hand side to get #8x+2=3x+17#.

Then subtract #3x# from both sides to get #5x+2=17#.

Then subtract #2# from both sides to get #5x=15#.

Then divide both sides by #5# to get #x=3#.

Plug back in to make sure #x=3# solves the equation.

#2(4*3+1)=26=26=3*3+17#, so it works.

Answer 2 · 2016-11-23T17:34:12

#x=3#

#2(4x+1)=3x+17#

1) multiply the brackets out

#8x+2=3x+17#

2) subtract#" " 2" " #from both sides

#8xcancel(+2-2)=3x+17-2#

#8x=+3x+15#

3) subtract 3x from both sides

#8x-3x=cancel(3x-3x)+15#

#5x=15#

4) divide both sides by 5

#(cancel(5)x)/cancel(5)=15/5=3#

#x=3#