How do you solve #4( x - 2) = x + 7#?

2 Answers
Jun 2, 2018

#x=5#

Explanation:

Given: #4(x-2)=x+7#.

Expand the left side.

#4x-8=x+7#

Subtract #7# from both sides.

#4x-8-7=x#

#4x-15=x#

Subtract #4x# from both sides.

#-15=x-4x#

#-15=-3x#

Divide by #-3#.

#x=(-15)/-3#

#=5#

Jun 2, 2018

#x=5#

Explanation:

To solve for the variable in the equation #4(x-2)=x+7#

Begin by eliminating the parenthesis by using the distributive property.

#4(x-2)=x+7#

#4x-8=x+7#

#3x-8=7#

Use the additive inverse to get the #x# terms on one side of the equation.

#4x-8-x= cancelx +7 cancel(-x)#

#3x-8=7#

Use the additive inverse to get the constant values on the same side of the equation

#3xcancel(-8) cancel(+8)=7+8#

#3x=15#

Now use the multiplicative inverse to solve for #x#

#(cancel(3)x)/cancel3 =15/3#

#x=5#