Question

HEEELPPPPP!!

Answer 1

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

Answer 2

what

Answer 3

hmm.. zero and negative exponents.. just simplify

Answer 4

teach me how to get the answer :)

Answer 5

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.

Answer 6

Do you still need help with simplifying?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

then divide 3x-2y .. ?

Answer 11

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

Answer 12

To multiply fractions, you multiply them across one another.

Answer 13

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

Answer 14

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

Answer 15

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 \[\frac{ -9 }{ 1 } (\frac{ 1 }{ y^2 })\].

Answer 16

x^2y^2

Answer 17

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

Answer 18

how ?

Answer 19

@izuru

Answer 20

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

Answer 21

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

Answer 22

?

Answer 23

@izuru

Answer 24

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.