## anonymous one year ago HEEELPPPPP!!

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1. anonymous

$\frac{ 4x^-2-9y^-2 }{ 3x-2y }$

2. anonymous

what

3. anonymous

hmm.. zero and negative exponents.. just simplify

4. anonymous

teach me how to get the answer :)

5. anonymous

Step one of simplifying is to use the negative power rule. The equation will be turned into $(4×\frac{ 1 }{ x^2}−9y×\frac{ 1 }{ y^2 })(3x−2y)$ Simplify even further by multiplying 4 and 9 with the fractions. Then, once you solved everything on the left parenthesis, cross multiply with the parenthesis on the right.

6. anonymous

Do you still need help with simplifying?

7. anonymous

hmm. $\frac{ 5x^2y^2+x^2y^2 }{ x^2y^2 } \times \frac{ 1 }{ 3x-2y }$

8. anonymous

Multiply $(4×\frac{1}{x^2})$ and $(−9y×\frac{1}{y^2})$

9. anonymous

= to 5x^2y^2+x^2y^2 over x^2y^2

10. anonymous

then divide 3x-2y .. ?

11. anonymous

$\frac{ 4}{ 1} (\frac{ 1 }{ x^2 })$ $\frac{ -9 }{ 1 } (\frac{ 1 }{ y^2 })$

12. anonymous

To multiply fractions, you multiply them across one another.

13. anonymous

Also, you don't divide 3x-2y. The equation calls to multiply it to the equation in parenthesis on the left.

14. anonymous

4x^2-9y^2+2 times 3x-2y ?

15. anonymous

The (4x^2−9y^2) part is correct. That is the numerator of the left parenthesis. Now, find the denominator. Look back on the denominators of $\frac{ 4}{ 1} (\frac{ 1 }{ x^2 })$ and $\frac{ -9 }{ 1 } (\frac{ 1 }{ y^2 })$.

16. anonymous

x^2y^2

17. anonymous

Correct. That's the denominator of the equation. Now, plug it in with the numerator. The equation is now $\frac{4{y}^{2}-9{x}^{2}}{{x}^{2}{y}^{2}}\times 3x-2y$. Cross multiply it

18. anonymous

how ?

19. anonymous

@izuru

20. anonymous

$\frac{4{y}^{2}-9{x}^{2}}{{x}^{2}{y}^{2}}\times \frac{ 3x-2y }{ 1 }$

21. anonymous

so the answer is $-\frac{ (3x+2y)(3x-2y)^2 }{ x^2y^2 }$

22. anonymous

?

23. anonymous

@izuru

24. anonymous

Okay so you cross multiply $\frac{4{y}^{2}-9{x}^{2}}{{x}^{2}{y}^{2}}\times 3x-2y$. The numerator can be simplified even further by using the difference of squares. The difference of squares fits the form a^2 - b^2. a = 2y and b = 3x. The equation rewritten will be $\frac{{(2y)}^{2}-{(3x)}^{2}}{{x}^{2}{y}^{2}}\times 3x-2y$. It can be simplified into $\frac{(2y+3x)(2y-3x)}{{x}^{2}{y}^{2}}\times 3x-2y$ Also, further simplification can be $\frac{3(2y+3x)(2y-3x)}{x{y}^{2}}-2y$ Moving the multiplication sign along with the squares.