# How do you solve 2p - ( 3- p ) \leq - 7p - 2?

Mar 21, 2017

$p \le \frac{1}{10}$

#### Explanation:

Given: $2 p - \left(3 - p\right) \le - 7 p - 2$

We need to get rid of the brackets and since there is a $-$ sign in front of the brackets, everything inside changes its sign:

$2 p - 3 + p \le - 7 p - 2$

Then we can sort out all the terms to move the unknowns from the numbers; be very careful with the signs:

$2 p - 3 + p \le - 7 p - 2$
$2 p + p + 7 p \le 3 - 2$
$10 p \le 1$

$p \le \frac{1}{10}$

That means if BIG $P$ were a centimeter $\left(c m\right)$ then our $p$ would be $\frac{1}{10} \textcolor{red}{\mathmr{and}}$ smaller than a millimeter $\left(m m\right)$.

To make sure the answer is correct, put it back into the $g i v e n$ equation. Now we are using the largest value that $p$ can be, or the value at which $p \textcolor{red}{=} \frac{1}{10}$.
We will then need to adjust the equation to reflect the following $e q u a l i t y$.

$2 p - \left(3 - p\right) \textcolor{red}{=} - 7 p - 2$

$2 \left(\frac{1}{10}\right) - \left(3 - \frac{1}{10}\right) \textcolor{red}{=} - 7 \times \frac{1}{10} - 2$

$2 \left(\frac{1}{10}\right) - \left(3 - \frac{1}{10}\right) \textcolor{red}{=} - 7 \times \frac{1}{10} - 2$

$\frac{2}{10} - 3 + \frac{1}{10} \textcolor{red}{=} - \frac{7}{10} - 2$

$- 2 - \frac{7}{10} \textcolor{red}{=} - \frac{7}{10} - 2$

Note: because $p \le \frac{1}{10}$ there are an infinite number of answers smaller than $p = \frac{1}{10}$ which when substituted into the $g i v e n$ equation will all result in an inequality.

Examples: $p \le \frac{1}{10}$ means:

p=1/10; p=1/50; p=-1/100; p=-100 ....