Question

"Solve the equation." Need help understanding how to do this! Pre cal semester review. Posting the equation in the comments so I can put it in the right form. Thank you!!

Answer 1

\[\frac{ -3x }{ x^2-4x-32 }-\frac{ 2 }{ x-8 }-\frac{ 3 }{ x+4 }\]

Answer 2

If you multiply the x-8 and the x+4 together, you get the quadratic expression in the first fraction. So in order to find the common denominator, you have to multiply the x-8 on the top and bottom by x+4. And then multiply the x+4 on top and bottom by x-8. Then every fraction will have a common denominator and you can combine the numerators over the common denominator.

Answer 3

Basic rules. Simply implement them with some idea how to proceed. You don't need to see the whole picture from the beginning. Just do SOMETHING that might seem helpful.

Adding fractions via the common denominator is almost always a good idea.

Answer 4

You are not solving an equation, by the way, because you don't have an equals sign in there at all. What you are doing is simply simplifying the expression.

Answer 5

Oh, I meant "=\[\frac{ 3 }{ x+4 }\] Oops! @IMStuck

Answer 6

@tkhunny Would I still do the s\[\frac{ -3x }{ x^2-4x-32 }-\frac{ 2 }{ x-8 }=\frac{ 3 }{ x+4 }\]

Answer 7

Do the same thing*

Answer 8

@3mar

Answer 9

Well, I am here.

Do you face any difficulties?

Answer 10

@3mar I need help with solving this problem.

Answer 11

I think @tkhunny is better than me at this...

But
I will not be late for any help.

Answer 12

@tkhunny Well, could you help me out? :)

Answer 13

Move the x-8 term to the other side.

Answer 14

Okay, so would I multiply both sides by (x-8)?

Answer 15

Well, you could do that, and that would get us somewhere, but that isn't what I meant. I suggestion adding 2/(x-8) to both sides.

Answer 16

Ah alright, done.

Answer 17

Okay, now find a common denominator on the right-hand side and add those two fractions. Show the result.

Answer 18

So a common denominator would be x+8, would I multiply 3/x+4 by 2 to get that? And would I have to multiply the left-hand side by 2 as well?

Answer 19

Seriously, we're just playing with it. This might lead us to a solution. If it doesn't, we'll try something else.

The common denominator is not (x-8). You have (x-8) and (x+4) to satisfy.

Try multiplying 2/(x-8) by (x+4)/(x+4) and then
multiplying 3/(x+4) by (x-8)/(x-8).

Just like adding numerical fractions with no common factor in the denominator.

1/3 + 1/7 = (1/3)*(7/7) + (1/7)*(3/3) = 7/21 + 3/21 = 10/21

Just like that.

Answer 20

Sorry the electricity was off.

Answer 21

\[x^2-4x-32=x^2-8x+4x-32=x(x-8)+4(x-8)=(x-8)(x+4 ) \]
\[\frac{ -3x }{ \left( x-8 \right)\left( x+4 \right) }-\frac{ 2 }{ x-8 }-\frac{ 3 }{ x+4 }\]
\[=\frac{ -3x-2\left( x+4 \right)-3\left( x-8 \right) }{ \left( x-8 \right)\left( x+4 \right) }\]
can you solve further?

Answer 22

@3mar I'm so confused...

Answer 23

Where are you stuck?

I want to simplify it to you......

Answer 24

If @tkhunny does not mind, I will proceed>>>

If that is our equation,
\[\Large\frac{ -3x }{ x^2-4x-32 }-\frac{ 2 }{ x-8 }=\frac{ 3 }{ x+4 }\]

you would better to transfer the term \(\frac{ 2 }{ x-8 }\) to the right-hand side:

\[\Large\frac{ -3x }{ x^2-4x-32 }=\frac{ 3 }{ x+4 }-\color{red}{(-\frac{ 2 }{ x-8 })}\]
\[\Large\frac{ -3x }{ x^2-4x-32 }=\frac{ 3 }{ x+4 }+\color{red}{\frac{ 2 }{ x-8 }}\]

Now, it is the turn to unify the denominators of the right-hand side:
\[\large \frac{ -3x }{ x^2-4x-32 }=\frac{ 3 }{ x+4 }+{\frac{ 2 }{ x-8 }}=\color{seagreen}{\frac{ 3(x-8)+2(x+4) }{ (x+4)(x-8) }}\]

Simplify the new numerator:
\[\large \frac{ -3x }{ x^2-4x-32 }={\frac{\color{lime}{ 3x-24+2x+8}}{ (x+4)(x-8) }}\]
\[\large \frac{ -3x }{ x^2-4x-32 }={\frac{\color{lime}{ 5x-16}}{ (x+4)(x-8) }}\]

...>>>... the chance to factor the denominator of the left-hand side fraction:
\[\LARGE x^2-4x-32=(\color{red}{x-8})(\color{red}{x+4})\]


and the new look of the equation will be:
\[\Huge\frac{ -3x }{ \color{blue}{(x-8)(x+4)} }={\frac{{ 5x-16}}{ \color{blue}{(x+4)(x-8) }}}\]

and you can notice from the last equation that the two denominators are equal (the same) and that leads to the equality of the numerators!

\[\LARGE \color{Hotpink}{-3x=5x-16}\]

Rearrage and simplify:
\[\LARGE \color{Hotpink}{-3x\color{red}{+3x}=5x-16\color{red}{+3x}}\]
\[\LARGE \color{Hotpink}{0=8x-16}\]
\[\LARGE \color{Hotpink}{0\color{red}{+16}=8x-16\color{red}{+16}}\]
\[\LARGE \color{Hotpink}{16=8x}\]
\[\LARGE \color{Hotpink}{\frac{ 16 }{ \color{red}8 }=\frac{ 8x }{ \color{red}8 }}\]

\[\Huge \bf\color{MediumSpringGreen }{x=2}\]

Answer 25

I hope I helped you.
I think you got the idea, clearer and satisfied!

I am waiting for your reply and comments if you have any! @sophiesky

*Sorry for being late for you!