# How do you solve 9c(c-11)+10(5c-3)=3c(c+5)+c(6c-3)-30?

Apr 18, 2017

#### Explanation:

First you want to distribute to all of the parentheses:

$9 {c}^{2} - 99 c + 50 c - 30 = 3 {c}^{2} + 15 c + 6 {c}^{2} - 3 c - 30$

Then you want to combine like factors.

$9 {c}^{2} - 49 c - 30 = 9 {c}^{2} + 12 c - 30$

Now that you have like factors on both sides, you can work to getting "c" by itself on one side.

$- 49 c = 12 c$

^^^However, the equation above is impossible because no matter what number you put (besides 0), you will not get the same answer from both sides.

For example:

$- 49 \left(1\right) = 12 \left(1\right)$
-49 is not equal to 12,
so impossible or undetermined.