Could someone explain how to do this problem?
Solve the following equation for u:
Simplify your answer as much as possible.
$\displaystyle \frac{3}{u+3} =-5$
I couldn't find how to do this in my notes.
You want to clear the fractions by multiplying by the LCD in this case
$\displaystyle (u+3)$
So mulitiplying both sides by the above gives...
$\displaystyle (u+3) \cdot \frac{3}{u+3} =-5 \cdot (u+3)$
This gives a new equation without any fractions
$\displaystyle 3=-5(u+3) \iff 3=-5u-15$
solving for u gives
$\displaystyle u=\frac{-18}{5}$
$\displaystyle \frac{3}{u+3} =-5$
Multiply both sides by $\displaystyle u+3$, the LCD
$\displaystyle \frac{3(u+3)}{u+3} =-5(u+3)$
$\displaystyle 3 =-5(u+3)$
$\displaystyle 3 =-5u-15$
Add 15 to both sides
$\displaystyle 3+15 =-5u-15+15$
$\displaystyle 18 =-5u$
Divide both sides by -5
$\displaystyle u =\frac{-18}{5}$
Okay, I think I've got it now.
I got a new problem with the same format. Could someone please check my answer to make sure I did it right?
$\displaystyle \frac{3}{x-1}=-2$
Multiplied both sides by (x-1)
$\displaystyle 3=-2x-2$
Add 2 to both sides
$\displaystyle 5=-2x$
And then divide both sides by -2.
$\displaystyle x=\frac{-5}{2}$
Did I do it right?