Question

Can someone help me with this Algebra 1 question?
Solve for x: -3|x + 7| = -12
and
Solve for x: −5|x + 1| = 10
I'll give you a medal and a fan if you answer both for me?

Answer 1

u can use the this theorem:
\[\large |a|=b\quad\Leftrightarrow\quad a=b\vee a=-b \]
given that \(b \geq0\).

Answer 2

I'm not sure what that is.

Answer 3

Or what it means

Answer 4

ok i will do one

Answer 5

\[\large -3\cdot|x+7|=-12 \]
\[\large |x+7|=\frac{-12}{-3}=4 \]
\[\large x+7=4\quad\text{or}\quad x+7=-4 \]
\[\large x=4-7=-3 \quad\text{or}\quad x=-4-7=-11 \]

Answer 6

essentially he's saying that instead of taking an absolute value as it is, you can split it up into two different situations. One where the term inside the absolute value bars equals the positive result, and one where it equals the negative result.

For example, with |x| = 3, x could equal positive or negative 3 and still satisfy the equation.
therefore
|x| = 3
can be rewritten without the absolute value bars as the two equations
x=3
and
x=-3

Answer 7

Thanks guys.

Answer 8

Wait so for the first one, would the answer be x=-3, x=1?