Solving equations.

**tim_mannire** · April 13th 2010, 05:59 PM

Hi, I need help with these. Simplfy as much as possible.

((3x

2^n+1) + (5x2^n)) / ((7x2^n) - (3x2^n-1))

solve for x

2y-1/1-x = 2y+1 x not =1 , y not =-1/2

solve for s

2s - 2/s = 3 s not = 0

**tim_mannire** · April 13th 2010, 06:01 PM

Originally Posted by tim_mannire

Hi, I need help with these. Simplfy as much as possible.

((3x2^n+1) + (5x2^n)) / ((7x2^n) - (3x2^n-1))

solve for x

2y-1/1-x = 2y+1 x not =1 , y not =-1/2

solve for s

2s - 2/s = 3 s not = 0

((3x is apart of the equation below.. i dont know y it isolated itself like that!

**dwsmith** · April 25th 2010, 11:45 AM

Wrong form but here is what you are looking for if this is the equation you mean

$\frac{3x2^{n+1}+5x2^n}{7x2^n-3x2^{n-1}}$

Factor out a $x$ and $2^n$

$\frac{x2^n(3*2+5)}{x2^n(7-\frac{3}{2})}=\frac{11}{\frac{11}{2}}=2$

Obtain the same common denominator

$\frac{2s^2-2}{s}=3\Rightarrow\frac{2(s^2-1)}{s}=3\Rightarrow\frac{s^2-1}{s}=\frac{3}{2}\Rightarrow s^2-1=\frac{3s}{2}\Rightarrow s^2-\frac{3s}{2}-1$ $=\frac{1}{2}(s-2)(2s+1)\Rightarrow s=2$ and $s=\frac{-1}{2}$

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