How do you solve #-(3x + 2) = -19x + 62#?

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Answer 1 · 2015-07-18T17:42:03

#x=4#

We're trying to get #x# alone. We need to find ways to manipulate the equation such that we can put all the #x#'s on one side of the equation.

Let's distrubute the negative outside the parenthesis first (you can equivalently divide by -1 on both sides of the equation)

#-(3x+2)=-3x-2=-19x+62#

Now, we can add the #19x# to both sides of the equation so that we're getting rid of it on the right side.

#-3x+19x-2=(-19x+19x)+62#

#16x-2=62#

We're close to getting #x# alone. We can add the #2# to both sides now so that it cancels out on the left side.

#16x=62+2=64#

Now, all we need to do is divide by #16# on both sides of the equation to get the value of #x#.

#x=64/16=4#