Elementary Algebra Question(29)

Algebra
Tutor: None Selected Time limit: 1 Day

Can I get help on the 29th question from this link please?

https://docs.google.com/file/d/0ByXp6Ibu-fCyOTBhMmE5MjgtN2U2Yy00YWI1LThmNDQtZjAzOWVmMTdmNjc1/edit?pli=1

May 18th, 2015

x/(x-6) - 3/(x+2) = 48/(x^2-4x-12)

[x*(x+2) - 3*(x-6)]/[(x-6)*(x+2)] = 48/(x^2-4x-12)

(x^2+2x-3x+18)/(x^2-4x-12) = 48/(x^2-4x-12)

(x^2-x+18)/(x^2-4x-12) = 48/(x^2-4x-12)

(x^2-x+18)/(x^2-4x-12) - 48/(x^2-4x-12) = 0

[(x^2-x+18)-48]/(x^2-4x-12)=0

(x^2-x-30)/(x^2-4x-12)=0

[(x-6)(x+5)]/[(x-6)(x+2)]=0

(x+5)/(x+2)=0

so x+5=0*(x+2)

x+5=0

x=-5

May 18th, 2015

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May 18th, 2015
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