Rational Equality Question

-Whiplash-

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Jan 22, 2014
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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?
 
If the only possible solution won't work, then there is no solution. ;)
 
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) \(\displaystyle \ \ \ \ \ \) <------ That is not the least common denominator. \(\displaystyle \ \)It's just a common denominator.
\(\displaystyle \ \) Look below the quote box.


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?


The least common denominator is the smallest product of the denominators such that each denominator divides it.


\(\displaystyle \displaystyle{\dfrac{1}{2a - 2} \ = \ \dfrac{a}{a^2 - 1} \ + \ \dfrac{2}{a + 1}}\)


\(\displaystyle \displaystyle{\dfrac{1}{2(a - 1)} \ = \ \dfrac{a}{(a - 1)(a + 1)} \ + \ \dfrac{2}{a + 1}}\)


The LCD = 2(a - 1)(a + 1).


\(\displaystyle \displaystyle{2(a - 1)(a + 1)\bigg[\dfrac{1}{2(a - 1)}\bigg] \ = \ 2(a - 1)(a + 1)\bigg[\dfrac{a}{(a - 1)(a + 1)}\bigg] \ + \ 2(a - 1)(a + 1)\bigg[\dfrac{2}{a + 1}}\bigg]\)


You don't get a quadratic equation:


\(\displaystyle a + 1 \ = 2a \ + \ 4(a - 1)\)


\(\displaystyle a + 1 \ = \ 6a - 4\)


\(\displaystyle 5 \ = \ 5a\)


\(\displaystyle a = 1, \ \ but \ \ it \ \ does \ \ not \ \ check.\)


\(\displaystyle \boxed{ \ No \ \ solution \ }\)
 
Last edited:
So I didn't notice that 2a-2 =2(a-1) would've saved me some work but I came to the same conclusion. there is no solution. thank you.
 
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