SOLUTION: Could someone explain to me how to solve: {{{ (3)/(x+2)+(1)/(x-2)=(7)/(x^2-4) }}} I get to: {{{ ((3)(x-2)+(1)(x+2))/(x+2)(x-2)=(7)/(x^2-4) }}} Then: {{{ (4x-4)/(x^2-4) = 7/(x^

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: Could someone explain to me how to solve: {{{ (3)/(x+2)+(1)/(x-2)=(7)/(x^2-4) }}} I get to: {{{ ((3)(x-2)+(1)(x+2))/(x+2)(x-2)=(7)/(x^2-4) }}} Then: {{{ (4x-4)/(x^2-4) = 7/(x^      Log On


   

Question 817753: Could someone explain to me how to solve:
+%283%29%2F%28x%2B2%29%2B%281%29%2F%28x-2%29=%287%29%2F%28x%5E2-4%29+
I get to: +%28%283%29%28x-2%29%2B%281%29%28x%2B2%29%29%2F%28x%2B2%29%28x-2%29=%287%29%2F%28x%5E2-4%29+
Then: +%284x-4%29%2F%28x%5E2-4%29+=+7%2F%28x%5E2-4%29+
And: +%284x-4%29%28x%5E2-4%29+=+7%28x%5E2-4%29+
Expand: +4x%5E3-16x-4x%5E2%2B16+=+7x%5E2-28+
Rearrange: +4x%5E3+-11x%5E2-16x%2B44+=+0+
Now I either made a serious mistake somewhere or maybe I just don't know what to do next.
Equation should have 2 answers
Thank you
Found 2 solutions by KMST, Edwin McCravy:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
+%284x-4%29%2F%28x%5E2-4%29+=+7%2F%28x%5E2-4%29+
At this point, keeping in mind that x=2 and x=-2 cannot be the final solutions, because they make denominators zero,
you just solve
4x-4=7
4x=11
x=11%2F4

NOTES:
The denominators in +%284x-4%29%2F%28x%5E2-4%29+=+7%2F%28x%5E2-4%29+
are the same on both sides of the equal sign,
so the numerators have to be the same.
You could say that you are multiplying both sides of the equation times x%5E2-4
to get an equation that is equivalent as long as x%5E2-4%3C%3E0 .
If you had found x=2 and/or x=-2 as solutions of the transformed equation,
you would say it is/they are "extraneous solution(s)",
solutions that were introduced by multiplying times x%5E2-4 that are not solutions of the original equation.

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
You did nothing mathematically wrong. The lady is correct.
However, you should clear of fractions from the beginning, 
not work on each side separately and then cross-multiply 
later.  It just happened to turn out that the denominators
on both sides were x²-4, but in other problems that will not
necessarily be the case.  So clear of denominators.  You
can probably skip the step where the red multiplication
and cancelling is below, because it's just a matter or
multiplying each numerator by the factors in the LCD that
it does not have underneath it.  You will be able to skip
down to 3(x-2)+1(x+2) = 7, just by multiplying each numerator
by the factors of the LCD which each numerator does not have
underneath it.  I put in the extra steps to show what you are
actually doing.

----------------------------------------------

+%283%29%2F%28x%2B2%29%2B%281%29%2F%28x-2%29=%287%29%2F%28x%5E2-4%29+

Factor {x²-4) as (x-2)(x+2)

+%283%29%2F%28x%2B2%29%2B%281%29%2F%28x-2%29=%287%29%2F%28%28x-2%29%28x%2B2%29%29+

Then get a least common denominator of
(x-2)(x+2) written as %28x-2%29%28x%2B2%29%2F%281%29

Then multiply every term by red%28%28x-2%29%28x%2B2%29%2F1%29



Then cancel



3(x-2)+1(x+2) = 7     <--- you could skip the above two steps to here
                           by multiplying each numerator by the factors
                           in the LCD which are not underneath it.

3x-6+x+2 = 7
4x-4 = 7
4x = 11
 x = 11%2F4

Also there is only one solution.

Edwin