A few days ago
Anonymous

Rational Equations?

I need a little bit of help solving these two rational equations:

(If you’re not able to see the whole problem, I think you can just move your cursor over the problem)

(x/2x-6)-(3/x^2-6x+9)=(x-2/3x-9)

and..

(2x+3/x-1)=(10/x^2-1)+(2x-3/x+1)

Top 2 Answers
A few days ago
MsMath

Favorite Answer

x/(2x-6) – 3/(x^2-6x+9) = (x-2) / (3x-9)

Factor each denominator.

2x – 6 = 2(x-3)

x^2 -6x +9 = (x-3)(x-3) = (x-3)^2

3x-9 = 3(x-3)

Multiply everything by 6(x-3)^2

3x(x-3) – 3(6) = (x-2)(2)(x-3)

Simplify

3x^2 – 9x – 18 = 2(x^2 -5x +6)

3x^2 -9x -18 = 2x^2 -10x + 12

Move everything to one side

x^2 + x – 30 = 0

Factor

(x+6)(x-5) = 0

x = -6 or x = 5

Follow the above steps for the second one.

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A few days ago
historian
First, remove the parentheses, changing the sign of what’s inside if the whole expression in parentheses has a – sign in front of it.

Then group the like terms and simplify them; if there are two or more terms that are the same power of x and have the same denominator

Then find common denominators for the fractions and combine.

Then see if you can find a way to factor any part of the equation. If you have a quadratic equation somewhere that can be factored, do it after isolating it on one side of the equation.

This kind of problem has to be chipped away at, little by little.

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