# Thread: Solve for the given variable:

1. ## Solve for the given variable:

I'm having trouble with these problems. Every time I see a fraction I get lost in these kind of equations.

Here is my first equation:

3x+4/x-2 = 5/6

and here is the second one:

8t - 6t-1/2 = 5

EDIT:
I apologize, I wrote my problems wrong!

This how they should be:

$\displaystyle \frac{3x+4}{x-2} = \frac{5}{6}$

And this is the second one:

$\displaystyle 8t-\frac{6t-1}{2} = 5$

Sorry, thanks for the help in advance!

2. $\displaystyle 3x+\frac{4}{x}-2 = \frac{5}{6}$

Multiply both sides through by x

$\displaystyle 3x^2+4-2x = \frac{5x}{6}$

That removed the fraction part in the second term.

Now multiply through by 6

$\displaystyle 18x^2+24-12x = 5x$

$\displaystyle 18x^2+24-17x = 0$

now you have a quadratic, can you solve this?

3. Thanks for the quick reply but I think you misunderstood my problem. I apologize, I wrote it incorrectly.
This is how it should look:

$\displaystyle \frac{3x+4}{x-2} = \frac{5}{6}$

And this is the second one:

$\displaystyle 8t-\frac{6t-1}{2} = 5$

I apologize, thank you for the help though.

4. Originally Posted by Somatic
I'm having trouble with these problems. Every time I see a fraction I get lost in these kind of equations.

Here is my first equation:

3x+4/x-2 = 5/6

and here is the second one:

8t - 6t-1/2 = 5

Fractions are our friends!

Just think of them as numbers. Look, if you don't like the fractions, replace it with a letter until the rest of the problem is done. Then, when you solve for x, put the fraction where it's supposed to go. It's like a crutch until you get it.

Check out the second problem.

if I pretend that 1/2 is $\displaystyle (a)$ for a minute, the I have

$\displaystyle 8t-6t-(a)=5$

Now, we want to get t by itself, so notice that $\displaystyle 8t$ and $\displaystyle 6t$ are like terms, so we can combine them like this

$\displaystyle 2t-(a)=5$

Now let's add $\displaystyle (a)$ to both sides

$\displaystyle 2t=5+(a)$

Now we divide both sides by two, and we have t alone at last

$\displaystyle t=\frac{5+(a)}{2}$

Now would be a good time to put the fraction back where $\displaystyle (a)$ is

$\displaystyle t=\frac{5+1/2}{2}$

Now, simplify the numerator

$\displaystyle \frac{5+1/2}{2}=\frac{\frac{10}{2}+\frac{1}{2}}{2}=\frac{\fr ac{11}{2}}{2}$

Now, I know this look crazy, but if you think about it

$\displaystyle \frac{\frac{11}{2}}{2}=\frac{11}{2}\div\frac{2}{1} =\frac{11}{2}\cdot\frac{1}{2}=\frac{11}{4}$

Therefore

$\displaystyle t=\frac{11}{4}$

This may confuse the heck out of you, but its another way to do things.

Bye!

5. Sorry guy!
I edited my first post, they have the correct equation. I wasn't sure on how to use the math tags. I'm sorry if I confused you all.
Thanks for the help though, I appreciate it!

Can someone give me a start on my problems please, thank you for the above posters for the help either way.

6. $\displaystyle \frac{3x+4}{x-2} = \frac{5}{6}$

$\displaystyle 6(3x+4) = 5(x-2)$

Now expand and group like terms.

7. Wow, thank you pickslides. I tend to over think things when I don't know how to begin these types of equations. I got my solution to the first problem which is:

$\displaystyle x = \frac{-34}{13}$

I am trying to solve the next one but I think I'm doing a step wrong.

$\displaystyle 8t - \frac{6t-1}{2} = 5$

I canceled the 2 by multiplying then I multiplied the other numbers as well.
$\displaystyle 16t-6t+1 = 10$

The 1 changes to a positive correct?

Am I on the right track?

EDIT:

I solved it.
I continued and did this.
$\displaystyle 10t+1 = 10$

End result :
$\displaystyle t = \frac{9}{10}$

Now I have another problem that really throws me off.

This:
$\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

Just give me a step please I want to try to solve but I don't know where to start.
Thanks for the support!

8. Originally Posted by Somatic

I am trying to solve the next one but I think I'm doing a step wrong.

$\displaystyle 8t - \frac{6t-1}{2} = 5$

I canceled the 2 by multiplying then I multiplied the other numbers as well.
$\displaystyle 16t-6t+1 = 10$

The 1 changes to a positive correct?

Am I on the right track?

EDIT:

I solved it.
I continued and did this.
$\displaystyle 10t+1 = 10$

End result :
$\displaystyle t = \frac{9}{10}$
Sounds good to me

9. $\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

multiply both sides by x gives

$\displaystyle 4 - \frac{x^2}{8} = 0$

Can you solve this?

10. Originally Posted by pickslides
$\displaystyle \frac{4}{x} - \frac{x}{8} = 0$

multiply both sides by x gives

$\displaystyle 4 - \frac{x^2}{8} = 0$

Can you solve this?
Okay, I'm not sure if I'm doing this correctly.

I then multiplied both sides by 8 giving me this:

$\displaystyle 4 - x^2 = 0$

After this continued with adding 4 to both sides got this:

$\displaystyle -x^2 = 4$

Is this my answer? I'm pretty sure I'm off big time. Please elaborate on what I did wrong, if I am incorrect.

$\displaystyle x = - \sqrt(4)$

11. You had the right idea, but didn't execute it too well.

$\displaystyle 4 - \frac{x^2}{8} = 0$

both sides by 8 gives

$\displaystyle 32 - x^2 = 0$

then

$\displaystyle 32 = x^2$

$\displaystyle x^2 = 32$

$\displaystyle x = \pm\sqrt{32}$