SOLUTION: Please help me solve: 2/5x^2-5x - 5/5x-5 r^2-25/(r+5)^2 x-6/x+6 = 5/7 d+1/d+4 divide by 3d+3/d-7 9/z-5 +21/(z-5)^2 Thank you, Gina

Question 449547: Please help me solve: 2/5x^2-5x - 5/5x-5
r^2-25/(r+5)^2
x-6/x+6 = 5/7
d+1/d+4 divide by 3d+3/d-7
9/z-5 +21/(z-5)^2
Thank you, Gina

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2/5x^2-5x - 5/5x-5
---
Factor:
[2/5x(x-1)] - 5/[5(x-1)]
-----
lcd = 5x(x-1)
---
Rewrite each fraction with the lcd as its denominator:
[2/lcd] - [5x/lcd]
----
Combine the numerators:
[2-5x]/lcd
----
= [2-5x]/[5x(x-1)]
======================

r^2-25/(r+5)^2
Factor:
[(r-5)(r+5)]/[(r+5)^2]
---
Cancel the common (r+5) to get:
= (r-5)/(r+5)
=======================
x-6/x+6 = 5/7
---
Cross-multiply:
7x-42 = 5x+30
2x = 72
x = 36
=========================
d+1/d+4 divide by 3d+3/d-7
---
Invert the denominator and multiply; also factor where you can:
[(d+1)/(d+4)]
---
Cancel the common d+1 factors:
---
= [1/(d+4)]
----
= (d-7)/[3(d+4)]
=====================
9/z-5 +21/(z-5)^2
----
lcd = (z-5)^2
---
Rewrite each fraction with the lcd
---
[9/lcd] + [21(z-5)]/lcd]
---
Combine the numerators over the lcd:
= [9+21z-105]/lcd
------------
= [21z--96]/[z-5]^2
=====================
Cheers,
Stan H.
==============

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