M mmonge New member Joined Oct 31, 2010 Messages 1 Oct 31, 2010 #1 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
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
D Denis Senior Member Joined Feb 17, 2004 Messages 1,700 Oct 31, 2010 #2 Post problems EXACTLY as they appear in your text book; what you posted makes NO SENSE.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 1, 2010 #3 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... \(\displaystyle \text{We have: }\;\frac{2}{x(x-1)} + \frac{8}{(x-1)(x+1)}\) \(\displaystyle \text{Get a common denominator:}\) . . \(\displaystyle \frac{2}{x(x-1)}\cdot\frac{x+1}{x+1} + \frac{8}{(x-1)(x+1)}\cdot\frac{x}{x} \;\;=\;\;\frac{2(x+1)}{x(x-1)(x+2)} + \frac{8x}{x(x-1)(x+1)}\) . . . . \(\displaystyle \frac{2(x+1) + 8x}{x(x-1)(x+1)} \;\;=\;\; \frac{10x+2}{x(x-1)(x+1)}\)
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... \(\displaystyle \text{We have: }\;\frac{2}{x(x-1)} + \frac{8}{(x-1)(x+1)}\) \(\displaystyle \text{Get a common denominator:}\) . . \(\displaystyle \frac{2}{x(x-1)}\cdot\frac{x+1}{x+1} + \frac{8}{(x-1)(x+1)}\cdot\frac{x}{x} \;\;=\;\;\frac{2(x+1)}{x(x-1)(x+2)} + \frac{8x}{x(x-1)(x+1)}\) . . . . \(\displaystyle \frac{2(x+1) + 8x}{x(x-1)(x+1)} \;\;=\;\; \frac{10x+2}{x(x-1)(x+1)}\)
