# Math Help - solving this fraction

1. ## solving this fraction

hi,
i want to know if im doing this fraction correctly? the question is at the top of the page.

after the last part i would do a quadratic equation.

would this be correct??

if not where am i going wrong??

thanks!

2. You correctly multiply each numerator by what's needed to get a common denominator, but then you ignore the common denominator! Since the RHS is not zero, your denominator matters in this problem.

3. i don't understand. which part on my link is where im going wrong??

4. Going from your first line to your second line, you have this:

$\dfrac{x+5}{3}-\dfrac{x-4}{2}=5$

$\dfrac{2x+10-3x+12}{x^{2}+x-20}=5.$

$\dfrac{2x+10-3x+12}{6}=5.$

5. oh i understand! sorry about that dunno why i put that!!!

so from there i would do:

6. Where did the 10 go from the first line to the second?

7. You should have $10 + 12 = 22$ in the numerator whereas you have 12. The method is fine though

$22-x = 30$

Remember you can (and should) always check the value of x obtained but subbing it into the original equation and seeing if it works

8. 22? wouldn't it be -2?? 10-12=-2

9. No: if you look at my post # 4, you will see that the 10 and the 12 have the same sign. Therefore, they must add up.

10. oh sorry i was looking at the minus from the qoriginal question!

so would i do what i did in post #5 but make it
-x+22=30
-x-8=0
-8/-1=8???

11. You're almost there. To solve for x, you get x all by itself on one side of the equation:

-x+22=30
-x=30-22=8
-x=8

Then what?

12. 8/-1= -8??

13. I would probably write it this way:

-x=8
-(-x)=-8
x=-8.

You want to have an equation at the end that says what you want to say. Presentation is important!

Does this answer satisfy the original equation?

14. so do i have answers? x=8 x=-8?

15. Here's a useful trick: when solving algebraic equations, you never need to check an answer key, because it's quite easy to check the answer yourself. All you have to do is plug your answer into the original equation, and see if it satisfies the equality. So what do you get when you plug your answer into the original equation?

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