1. ## Can you check my answer?

We are asked to solve: $f(x)=\frac{4x+8}{x-3}$

I know this is simple, but im reallly rusty. So what I tried to do is get rid of the x-3 by multiply both sides by it

$\frac{4x+8}{x-3}=0$ = $x-3=4x+8$

Then I moved the variables to one side and the integers to another:

$x-4x=3+8$
$-3x=11$

Then I divided both sides by -3
$x=\frac{11}{-3}$

2. ## Re: Can you check my answer?

Originally Posted by NewGuy21
We are asked to solve: $f(x)=\frac{4x+8}{x-3}$

I know this is simple, but im reallly rusty. So what I tried to do is get rid of the x-3 by multiply both sides by it

$\frac{4x+8}{x-3}=0$ = $x-3=4x+8$

Then I moved the variables to one side and the integers to another:

$x-4x=3+8$
$-3x=11$

Then I divided both sides by -3
$x=\frac{11}{-3}$

First the question: are you trying to solve $\displaystyle f(x) = 0$? If so then ...

The first line where you multiply top and bottom by $\displaystyle x - 3$ note that $\displaystyle (x - 3) \times 0 \ne x - 3$.

Carry on from there ...

3. ## Re: Can you check my answer?

Just to add - what you solved for was f(x)=1, not f(x) =0. Some advice: always double-check your work, in this case by substituting x=-11/3 back into the original equation to see if it works out. If you put x=-11/3 into the equation you get:

$\displaystyle f(-11/3) = \frac {4(\frac {-11} 3 ) +8}{\frac {-11} 3 -3} = \frac { \frac {-44+24} 3}{\frac {-11-9}{3}} = \frac {-20}{-20} = 1$

4. ## Re: Can you check my answer?

Originally Posted by NewGuy21
We are asked to solve: $f(x)=\frac{4x+8}{x-3}$

I know this is simple, but im reallly rusty. So what I tried to do is get rid of the x-3 by multiply both sides by it

$\frac{4x+8}{x-3}=0$ = $x-3=4x+8$

Then I moved the variables to one side and the integers to another:

$x-4x=3+8$
$-3x=11$

Then I divided both sides by -3
$x=\frac{11}{-3}$