anonymous
  • anonymous
HEEELPPPPP!!
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\frac{ 4x^-2-9y^-2 }{ 3x-2y }\]
anonymous
  • anonymous
what
anonymous
  • anonymous
hmm.. zero and negative exponents.. just simplify

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anonymous
  • anonymous
teach me how to get the answer :)
anonymous
  • 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.
anonymous
  • anonymous
Do you still need help with simplifying?
anonymous
  • anonymous
hmm. \[\frac{ 5x^2y^2+x^2y^2 }{ x^2y^2 } \times \frac{ 1 }{ 3x-2y }\]
anonymous
  • anonymous
Multiply \[(4×\frac{1}{x^2})\] and \[(−9y×\frac{1}{y^2})\]
anonymous
  • anonymous
= to 5x^2y^2+x^2y^2 over x^2y^2
anonymous
  • anonymous
then divide 3x-2y .. ?
anonymous
  • anonymous
\[\frac{ 4}{ 1} (\frac{ 1 }{ x^2 })\] \[\frac{ -9 }{ 1 } (\frac{ 1 }{ y^2 })\]
anonymous
  • anonymous
To multiply fractions, you multiply them across one another.
anonymous
  • anonymous
Also, you don't divide 3x-2y. The equation calls to multiply it to the equation in parenthesis on the left.
anonymous
  • anonymous
4x^2-9y^2+2 times 3x-2y ?
anonymous
  • 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 \[\frac{ -9 }{ 1 } (\frac{ 1 }{ y^2 })\].
anonymous
  • anonymous
x^2y^2
anonymous
  • 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
anonymous
  • anonymous
how ?
anonymous
  • anonymous
@izuru
anonymous
  • anonymous
\[\frac{4{y}^{2}-9{x}^{2}}{{x}^{2}{y}^{2}}\times \frac{ 3x-2y }{ 1 }\]
anonymous
  • anonymous
so the answer is \[-\frac{ (3x+2y)(3x-2y)^2 }{ x^2y^2 }\]
anonymous
  • anonymous
?
anonymous
  • anonymous
@izuru
anonymous
  • 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.

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