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

Question

2111 views around the world

You can reuse this answer
Creative Commons License

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 $x$ .

$x=64/16=4$

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