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Question about rational equations
- Thread starter user12341
- Start date
You add algebraic fractions the exact same way you add numeric functions.(-2x)/(x-1)+(x+3)/(x^2 - 1)=1
I do not understand how you combine the left side into one fraction.
\(\displaystyle \dfrac{2}{3} + \dfrac{2}{7} = \dfrac{2 * 7}{3 * 7} + \dfrac{2 * 3}{7 * 3} = \dfrac{14}{21} + \dfrac{6}{21} = \dfrac{20}{21}.\)
I turned each fraction into an equal fraction such that both fractions had a common denominator.
To make life easy, you probably were to told to turn fractions into equivalents using the least common denominator. Ring a bell?
\(\displaystyle \dfrac{5}{6} + \dfrac{1}{3} = \dfrac{5}{6} + \dfrac{2}{6} = \dfrac{7}{6} = 1 + \dfrac{1}{6}.\)
In the example above you only needed to convert one fraction because the denominator of one is a multiple of the other.
Things are exactly the same in algebra because the expressions just represent ordinary numbers.
Now notice that \(\displaystyle x^2 - 1 = (x - 1)(x + 1).\) So one denominator is a multiple of the other. So you can leave one fraction alone. Which one?
What do you do to the numerator and denominator of the other fraction to get an equivalent with the least common denominator?
HOWEVER, you really do want fractions at all. GET RID of them by multiplying the whole equation by the least common denominator. No one likes working with fractions. So make life easy and get rid of them as soon as possible.
D
Deleted member 4993
Guest
My problem is that i end up with some x^2's towards the end and I know the answer is supposed to be -4/3. I don't know where my mistake is, could you walk me through getting it to a denominator of x+1 and/or solving it.
If you walk us through the steps of getting "....some x^2's towards the end" - we can catch your mistake and correct.
That would be the preferred way for us ...
I have tried several variations but this was the last one i tried:
(-2x(x+1))/(x^2-1) + (x+3)/(x^2-1) = 1
(-2x^2 - 2x + x + 3)/(x^-1) = 1
-2x^2 - x + 3 = x^2 -1
-3x^2 - x = -4
????
but yet, when i look it up on a calculator it says i should get a fraction with the denominator x+1 by simplifying (adding) those 2 fractions together?
(-2x(x+1))/(x^2-1) + (x+3)/(x^2-1) = 1
(-2x^2 - 2x + x + 3)/(x^-1) = 1
-2x^2 - x + 3 = x^2 -1
-3x^2 - x = -4
????
but yet, when i look it up on a calculator it says i should get a fraction with the denominator x+1 by simplifying (adding) those 2 fractions together?
D
Deleted member 4993
Guest
.I have tried several variations but this was the last one i tried:
(-2x(x+1))/(x^2-1) + (x+3)/(x^2-1) = 1
(-2x^2 - 2x + x + 3)/(x^-1) = 1
-(2x+3)(x-1)/[(x+1)(x-1)] =1
-(2x+3)/(x+1) = 1
and continue......
-2x^2 - x + 3 = x^2 -1
-3x^2 - x = -4
????
but yet, when i look it up on a calculator it says i should get a fraction with the denominator x+1 by simplifying (adding) those 2 fractions together?
but that would give me x = 2/3, when i know for a fact that the answer should be -4/3.-(2x+3)(x-1)/[(x+1)(x-1)] =1
-(2x+3)/(x+1) = 1
and continue......
but that would give me x = 2/3, when i know for a fact that the answer should be -4/3
user12341, what you must have mistakenly done was:
-(2x + 3)/(x + 1) = 1
-2x + 3 = x + 1 . . . . . failed to distribute the negative sign to the "3."
3 - 1 = x + 2x
2 = 3x
2/3 = (3x)/3
2/3 = x . . . . . . This is an incorrect answer to the original problem.
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Instead, if x is not equal to -1, then:
-(2x + 3)/(x + 1) = 1 ---->
-(2x + 3) = x + 1 ---->
-2x - 3 = x + 1 ---->
-2x - x - 3 + 3 = x - x + 1 + 3 ---->
-3x = 4 ---->
(-3x)/(-3) = 4/(-3) ---->
x = -4/3
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