-Whiplash-
New member
- Joined
- Jan 22, 2014
- Messages
- 24
Hi, I was given the following question:
Solve the equation. Check your Solution.
\(\displaystyle \displaystyle{\dfrac{1}{2a-2}=\dfrac{a}{a^2-1}+\dfrac{2}{a+1}}\)
So I get the LCD: (2a-2)(a+1)(a-1)
Start canceling:
\(\displaystyle \displaystyle{(2a-2)(a+1)(a-1)\dfrac{1}{2a-2}=(2a-2)(a+1)(a-1)\dfrac{a}{a^2-1}+(2a-2)(a+1)(a-1)\dfrac{2}{a+1}}\)
Am left with
a2-1 = a(2a-2) + 2(2a-2)(a-1) Which becomes:
a2-1 = 6a2-10+4 I made it equal to zero:
0 = 5a2-10+5
I did the quadratic formula and came up with a = 1 which is all well and good except a can't equal 1 because that would make the denominator in the original equation 0.
What do I do? is this question really unsolvable?
Solve the equation. Check your Solution.
\(\displaystyle \displaystyle{\dfrac{1}{2a-2}=\dfrac{a}{a^2-1}+\dfrac{2}{a+1}}\)
So I get the LCD: (2a-2)(a+1)(a-1)
Start canceling:
\(\displaystyle \displaystyle{(2a-2)(a+1)(a-1)\dfrac{1}{2a-2}=(2a-2)(a+1)(a-1)\dfrac{a}{a^2-1}+(2a-2)(a+1)(a-1)\dfrac{2}{a+1}}\)
Am left with
a2-1 = a(2a-2) + 2(2a-2)(a-1) Which becomes:
a2-1 = 6a2-10+4 I made it equal to zero:
0 = 5a2-10+5
I did the quadratic formula and came up with a = 1 which is all well and good except a can't equal 1 because that would make the denominator in the original equation 0.
What do I do? is this question really unsolvable?