+0

# pls show steps

+1
68
5

pls show steps

Feb 24, 2021

#1
+939
0

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

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

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

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

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

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

.
Feb 24, 2021
#2
+485
0

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

Logarhythm  Feb 24, 2021
#5
+939
+1

CubeyThePenguin  Feb 24, 2021
#3
+485
+2

Excuse the handwriting..

Feb 24, 2021
#4
+266
+2

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

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

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

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

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

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

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

Feb 24, 2021