# How do you solve -10+x+4-5=7x-5?

Aug 26, 2016

First, you gather all the "x" terms on one side of the equation;

#### Explanation:

and all of the constants on the other side of the equation.

You have:
$- 10 + x + 4 - 5 = 7 x - 5$

1st: subtract x from both sides of the equation.
(We want to move the x over to the right-hand side, so we do the opposite operation on x. X is positive on the left, so we subtract, and we have to do the same thing to both sides of an equation.):

$- 10 + x + 4 - 5 = 7 x - 5$
$\ldots \ldots \ldots - x \ldots \ldots \ldots . . - x$

You then have:
$- 10 + 4 - 5 = 6 x - 5$

All the x's are now on the right-hand side of the equation.

Now combine the constants on the left side of the equation:
$- 10 + 4 - 5 = - 15 + 4 = - 11$

so we have:
$- 11 = 6 x - 5$

There is one more constant to move, so add it to both sides:
We have: $- 11 + 5 = 6 x$ , which is $- 6 = 6 x$

Let's write it as:
$6 x = - 6$

Now we need to divide both sides by - 6 in order to end up with only one x:
$\frac{6 x}{-} 6 = - \frac{6}{-} 6$,

which equals:
$- x = 1$

Since we want to know what positive x is, we now multiply both sides of the equation by $- 1$ to get:

$x = - 1$

Now check by plugging $- 1$ back into the original equation:
$- 10 - 1 + 4 - 5 = 7 \left(- 1\right) - 5$

Do both sides add up to negative 12? They should.