Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!

I need help with 5 equations... this is one of them.

8/x+2 + 8/2 = 5

Please help me!!!

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- Jan 5th 2007, 11:41 AMmOoGaLySolving Rational Equations...help!!
Okay, I don't know how to do this what so ever and I'm getting frustrated because my book isn't showing in great detail what I'm suppose to be doing!

I need help with 5 equations... this is one of them.

8/x+2 + 8/2 = 5

Please help me!!! - Jan 5th 2007, 11:56 AMgalactus
You haven't used parentheses, so your post is rather ambiguous.

Is this it:

$\displaystyle \frac{8}{x+2}+\frac{8}{2}=5$

If so, why not write 8/2 as 4?.

Anyway, multiply by $\displaystyle x+2$:

$\displaystyle (x+2)\frac{8}{x+2}+4(x+2)=5(x+2)$

$\displaystyle 8+4(x+2)=5(x+2)$

Now, can you finish?. - Jan 5th 2007, 12:11 PMmOoGaLy
- Jan 5th 2007, 12:35 PMgalactus
No, check yourself. If you put your answer back into the equation, do you get 5?. I don't think so.

You got a little discombobulated on your distribution.

Let's try your answer first:

$\displaystyle \underbrace{\frac{8}{\underbrace{-2+2}_{\text{oops, division by 0, a big no-no}}}+4=5}_{\text{bad, the universe will\\collapse in on itself}}$

Now, let's do it this way:

$\displaystyle 8+4(x+2)=5(x+2)$

$\displaystyle 8+4x+8=5x+10$

$\displaystyle 16+4x=5x+10$

Subtract 4x from both sides:

$\displaystyle 16=x+10$

Subtract 10 from both sides:

$\displaystyle x=6$

See?.

Actually, you're not seeing the forest for the trees. Don't let the algebra and x's psyche you out.

Look at it for a second. Follow:

$\displaystyle \frac{8}{x+2}+4=5$

What plus 4 equals 5. Well, you don't have to be a brilliant mathematician to see that. 4+1=5

Therefore, $\displaystyle \frac{8}{x+2}$ must be 1. What makes that 1?. When x=6. See?.

Relax. It'll be OK. - Jan 5th 2007, 12:46 PMmOoGaLy
Ohh. Parenthasis first..

That seems so easy! Thank you!!

$\displaystyle \frac{2}{3x}+\frac{2}{3}=\frac{8}{x+6}$

Do I multiply $\displaystyle x+6$ by all the equations? - Jan 5th 2007, 01:23 PMgalactus
You're partially correct. You use the GCD, which is $\displaystyle 3x(x+6)$

See?. You have more in the denominators than just x+6. There's a 3x there, also.

$\displaystyle (x+6)(3x)\frac{2}{3x}+(x+6)(3x)\frac{2}{3}=(x+6)(3 x)\frac{8}{x+6}$

Cancel what needs cancelled and you get:

$\displaystyle 2(x+6)+2(x+6)(x)=8(3x)$

Distribute:

$\displaystyle 2x+12+2x^{2}+12x=24x$

$\displaystyle 14x+12+2x^{2}=24x$

$\displaystyle 2x^{2}-10x+12=0$

Divide by 2:

$\displaystyle x^{2}-5x+6=0$

Now, can you solve the quadratic?. - Jan 5th 2007, 01:42 PMmOoGaLy
Woah woah, why do you divide by 2?

Okay, well using the equation $\displaystyle x^{2}-5x+6=0$

I used the quadratic formula and I am stuck at

(doesn't know the codes to make this look correct)

x = -5 plus/minus symbol 1 all over 2 - Jan 5th 2007, 02:04 PMgalactus
I divided by 2 because I could and wanted to. All the terms were divisible by 2

The quadratic $\displaystyle x^{2}-5x+6=0$ can be factored.

Can you factor?. You can use the quadratic formula, too, but this one is easily factorable.

$\displaystyle (x-3)(x-2)=0$ - Jan 5th 2007, 02:07 PMmOoGaLy
- Jan 5th 2007, 02:47 PMmOoGaLy$\displaystyle \frac{x-3}{x-4}+4=\frac{3x}{x}$

$\displaystyle (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}$

I multipied them

$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)$ but I had a feeling that I was suppose to cancel out the 4's but I chose not to..

Next, $\displaystyle x^{2}-3x+4x^{2}-16x=3x^{2}-12x$

And I moved the $\displaystyle 3x^{2}-12x$

to the left side of the equation...

$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x$

$\displaystyle 5x^{2}-3x{2}-19x-12x$

I'm stuck at $\displaystyle 2x^{2}-21x$

I did something wrong, didn't I? =( I am horrible at this! - Jan 5th 2007, 03:01 PMgalactus
$\displaystyle (x-4)(x)\frac{x-3}{x-4}+(x-4)(x)4=(x-4)(x)\frac{3x}{x}$

I multipied them

$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)$ but I had a feeling that I was suppose to cancel out the 4's but I chose not to..

Next, $\displaystyle x^{2}-3x+4x^{2}-16x=3x^{2}-12x$

And I moved the $\displaystyle 3x^{2}-12x$

to the left side of the equation...

$\displaystyle x(x-3)+4(x-4)(x)=3x(x-4)=3x^{2}-12x$$\displaystyle \leftarrow\text{good to here}$

$\displaystyle 5x^{2}-19x=3x^{2}-12x$

$\displaystyle 2x^{2}-7x=0$

Factor:

$\displaystyle x(2x-7)=0$

See the solution now?.

$\displaystyle 5x^{2}-3x{2}-19x12x$

I'm stuck at $\displaystyle 2x^{2}-21x$

I did something wrong, didn't I? =( I am horrible at this! - Jan 5th 2007, 03:18 PMmOoGaLy
You lost me. How does $\displaystyle x(2x-7)=0$ end up into x = a number?

- Jan 5th 2007, 03:19 PMmOoGaLy
:confused: $\displaystyle 2x^{2}=7$ ??

- Jan 5th 2007, 06:08 PMtopsquark
$\displaystyle x(2x-7) = 0$

Since the RHS is 0, at least one of x or 2x - 7 must also be 0. Thus:

$\displaystyle x = 0$ or $\displaystyle 2x - 7 = 0$

Thus $\displaystyle x = 0$ or $\displaystyle x = \frac{7}{2}$

Now be careful. Your original equation canNOT work with x = 0. (The RHS of your original equation has us dividing 0/0, which is not allowed.) So your only solution is $\displaystyle x = \frac{7}{2}$.

-Dan - Jan 7th 2007, 02:37 PMmOoGaLy
Oh! Okay, thank you so much galactus & topsquark!! You've been loads of help.

If I need anymore, I'll be back lol