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pls show steps

 Feb 24, 2021
 #1
avatar+964 
0

\(\frac{y-2}{y^2 - 9} - \frac{y-7}{y^2 + 4y - 21}\)

 

\(\frac{y-2}{(y+3)(y-3)} - \frac{y-7}{(y+7)(y-3)}\)

 

\(\frac{(y+7)(y-2) - (y+3)(y-7)}{(y+3)(y-3)(y+7)}\)

 

\(\frac{(y^2 + 5y - 14) - (y^2 - 4y - 21)}{(y+3)(y-3)(y+7)}\)

 

\(\frac{y+7}{(y+3)(y-3)(y+7)}\)

 

\(\boxed{\frac{1}{(y+3)(y-3)}}\)

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 Feb 24, 2021
 #2
avatar+544 
0

Nice answer Cubey, but I think I got something different from yours..

Logarhythm  Feb 24, 2021
 #5
avatar+964 
+1

Logarhythm's answer is correct.

CubeyThePenguin  Feb 24, 2021
 #3
avatar+544 
+2

Excuse the handwriting..

 

 Feb 24, 2021
 #4
avatar+235 
+1

Hmmm, I got something similar to what Log got...

\(\frac{y-2}{y^2\:-\:9}\:-\:\frac{y-7}{y^2\:+\:4y\:-\:21}\)

 

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{y^2+4y-21}\)

 

\(=\frac{y-2}{\left(y+3\right)\left(y-3\right)}-\frac{y-7}{\left(y-3\right)\left(y+7\right)}\)

 

\(=\frac{\left(y-2\right)\left(y+7\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}-\frac{\left(y-7\right)\left(y+3\right)}{\left(y-3\right)\left(y+7\right)\left(y+3\right)}\)

 

\(=\frac{\left(y-2\right)\left(y+7\right)-\left(y-7\right)\left(y+3\right)}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

 

\(=\frac{9y+7}{\left(y+3\right)\left(y-3\right)\left(y+7\right)}\)

 

cheekycheekycheeky

 Feb 24, 2021

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