# How do i solve equations with negative numbers? including adding, subtracting, dividing, and multiplying?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

3

### This answer has been featured!

Featured answers represent the very best answers the Socratic community can create.

Feb 17, 2017

For a while, explicitly put the -1

#### Explanation:

Let us agree on this $- 5 = \left(5\right) \left(- 1\right)$

Anything multiplied by 1 is itself, and anything multiplied by (-1) is its opposite.

Let us agree on this $5 = \left(- 1\right) \left(- 1\right) \left(5\right)$

A positive number multiplied by (-1) results in that negative number, and a negative number multiplied by (-1) results in that positive number.

A positive number multiplied by (-1) an even number of times results in that positive number (no change).

A negative number multiplied by (-1) an even number of times results in that negative number (no change).

Multiplying by (-1) twice undoes the multiplication.

So, with this in mind, let us consider addition.

$- 2 + 3$

We can write this as

$\left(- 1\right) \left(2\right) + 3$

Addition has a property that allows us to do it in any order, and still get the same result. It's called the communitive property.

$a + b = b + a$

Well, we just turned our problem into an addition problem. That means we can rearrange the terms. So, let's do that:

$3 + \left(- 1\right) \left(2\right)$

According to order of operations, we must multiply before adding, so let's multiply that (-1):

$3 - 2$

So, it appears that an addition problem with a negative in front is really a subtraction problem in disguise.

Let's try another one:

$- 2 - 3$

We will again replace the negatives with (-1), but it is important to remember that we are adding. We always add, but sometimes we add negative numbers.

$\left(- 1\right) \left(2\right) + \left(- 1\right) \left(3\right)$

Both terms are being multiplied by (-1), which brings us to another property. The distributive property says:

$a b + a c = a \left(b + c\right)$

Let us pull out the (-1) in like fashion.

$\left(- 1\right) \left(2 + 3\right) = \left(- 1\right) \left(5\right) = - 5$

Finally, we have

$2 - 3$

Again, this can be thought of as:

$2 + \left(- 1\right) \left(3\right)$

Move the bigger number to the front

$\left(- 1\right) \left(3\right) + 2$

Multiply first

$- 3 + 2$

We can pull out a (-1) here too because anything multiplied by 1 is itself and anything multiplied by (-1) is its opposite, so positive becomes negative in that case.

$\left(- 1\right) \left(3 - 2\right)$

$= \left(- 1\right) \left(1\right)$

$= \left(- 1\right)$

Now, it would be silly to do all of this every time. You will very rapidly internalize these ideas, but hopefully this will help in thinking about it.

We have covered addition, multiplication, and subtraction. Division might be a little tricky, but I know you can get it.

$- \frac{2}{3} = \frac{- 2}{3} = \frac{2}{-} 3$

Let us see them more clearly:

$\left(- 1\right) \frac{2}{3} = \frac{\left(- 1\right) 2}{3} = \frac{2}{\left(- 1\right) 3}$

Remember that an even number of (-1) produces no change.

$\frac{2}{3} = \frac{\left(- 1\right) 2}{\left(- 1\right) 3} = \frac{\left(\cancel{- 1}\right) 2}{\left(\cancel{- 1}\right) 3}$

So, with this knowledge, let's solve an equation.

$\frac{- 5 + 5 - 2 + 7 \cdot - 2}{-} 2$

$= \frac{\left(- 1\right) 5 + 5 + \left(- 1\right) 2 + 7 \cdot \left(- 1\right) 2}{\left(- 1\right) 2}$

$= \frac{\left(- 1\right) \left(5 - 5\right) + \left(- 1\right) 2 + \left(- 1\right) 14}{\left(- 1\right) 2}$

$= \frac{\left(\cancel{- 1}\right) \left(0\right) + \left(\cancel{- 1}\right) 2 + \left(\cancel{- 1}\right) 14}{\left(\cancel{- 1}\right) 2}$
Note:All terms contain (-1), so it is a common factor.

$= \frac{\left(0\right) + 2 + 14}{2}$

$= \frac{2 + 14}{2}$

$= \frac{16}{2}$

$= 8$

Get the hang of it, and then abandon it. It will just be automatic.

• 15 minutes ago
• 18 minutes ago
• 19 minutes ago
• 19 minutes ago
• 10 minutes ago
• 10 minutes ago
• 11 minutes ago
• 12 minutes ago
• 13 minutes ago
• 13 minutes ago
• 15 minutes ago
• 18 minutes ago
• 19 minutes ago
• 19 minutes ago