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Mathematics 6 Online
OpenStudy (anonymous):

Solve: (x/(x-3))+ (9/(x+1))=4

OpenStudy (anonymous):

\[\frac{x}{x-3}+\frac{9}{x+1}=\frac{x(x+1)+9(x-3)}{(x-3)(x+1)}\] is a start

OpenStudy (anonymous):

you get \[\frac{x^2+x+9x-27}{(x-3)(x+1)}=\frac{x^2+10x-27}{(x-3)(x+1)}=4\]

OpenStudy (anonymous):

so \[x^2+10x-27=4(x-3)(x+1)\] and so multiply out and get a quadratic equation (which actually factors) you good from there?

OpenStudy (anonymous):

Yes, I believe so. Thanks again. You're a lifesaver.

OpenStudy (gw2011):

x/(x-3)+9/(x+1)=4 Multiply by x-3 to eliminate the denominator x-3 x+(9)(x-3)/(x+1)=4(x-3) Multiply by x+1 to eliminate the denominator x+1 x(x+1)+9(x-#)=4(x-3)(x+1) x^2+x+9x-27=4(x^2-2x-3) x^2+10x-27=4x^-8x-12 4x^2-8x-12-x^2-10x+27=0 3x^-18x+15=0 (3x-3)(x-5)=0 3x-3=0 3x=3 x=1 x-5=0 x=5 Therefore, x=1 and x=5

OpenStudy (anonymous):

Yes, I believe so. Thanks again. You're a lifesaver.

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