Rational Expressions

M

mmonge

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Oct 31, 2010
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I am stuck on these two problems i would greatly appreciate some help...

1. Y= U+1 SOLVE FOR U
U-1


The other one is
2. 2 8
x^2-x + x^2-1
 
Post problems EXACTLY as they appear in your text book; what you posted makes NO SENSE.
 
Hello, mmonge!

\(\displaystyle \text{1. Solve for }u\!:\;\; y \:=\:\frac{u+1}{u-1}\)
Click to expand...

\(\displaystyle \text{Multiply by }(u-1)\!:\;\;y(u-1) \;=\;u+1 \quad\Rightarrow\quad uy - y \;=\;u + 1\)

. . . . . . . . . . . . . . . .\(\displaystyle uy - u \;=\;y + 1 \quad\Rightarrow\quad u(y-1) \;=\;y+1\)

\(\displaystyle \text{Therefore: }\;u \;=\;\frac{y+1}{y-1}\)



\(\displaystyle \text{2. Simplify: }\;\; \frac{2}{x^2-x} + \frac{8}{x^2-1}\)
Click to expand...

\(\displaystyle \text{We have: }\;\frac{2}{x(x-1)} + \frac{8}{(x-1)(x+1)}\)


\(\displaystyle \text{Get a common denominator:}\)

. . \(\displaystyle \frac{2}{x(x-1)}\cdot\frac{x+1}{x+1} + \frac{8}{(x-1)(x+1)}\cdot\frac{x}{x} \;\;=\;\;\frac{2(x+1)}{x(x-1)(x+2)} + \frac{8x}{x(x-1)(x+1)}\)

. . . . \(\displaystyle \frac{2(x+1) + 8x}{x(x-1)(x+1)} \;\;=\;\; \frac{10x+2}{x(x-1)(x+1)}\)

 
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