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Question 251968: I have to solve for 'z'
z/(z-2) - 1/(z+5) = 7/z^2+3z-10
Now I know that I have to find the common denominator which would be z^2+3z-10, now my question is do I multiply the first denominator by z+3z+5 and the second one by z+3z-2
If I did that I would have z/z^2+3z-10 - 1/z^2++3z-10 = 7/z^2+3z-10
z/z^2+3z - 1/z^2+3z-10 = 7/z^2+3z-10
Am I doing this correctly, or did I get lost?
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! I interpret it as
z/(z-2) - 1/(z+5) = 7/(z^2+3z-10 )
Where there are three denominators
(z-2) , (z+5) , and (z^2+3z-10 )
the problem is simpler than you imagine because (z^2+3z-10 )can be factored into (z-2)(z+5)
and the solution is z=1
So if we multiply both sides by (z-2)(z+5)
we get
z*(z+5)-1(z-2)=7
z^2+5z-z+2=7
z^2+4z=5
so now we complete the square
add 4 to both sides 1/2*4=2 and square it =4
z^2+4z+4=9
(z+2)^2=9
z+2=+\-3
z=-2+3=1
z=-2-3=-5
if you go back in the original equations you will find that -5 doesn't work because it cause a zero denominator which is undefined
so the only solution is z=1
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