Question

what is the solution to the rational equation 5/x^2-3x+2 - 1/x-2 = 1/3x-3

Answer 1

hmm u need brackets so that i can see the equation exactly and not make mistakes

Answer 2

@Hero

Answer 3

Use the equation editor to re-post your question. For example, to make the fraction \[\frac{a}{b}\] type frac{a}{b} in equation editor.

Answer 4

okay, did you not get the picture though?

Answer 5

@Hero

Answer 6

Oh, okay, I see the picture now.

Answer 7

The way to do it is to substitute values for x into the equation one at a time until you get a solution that works.

Answer 8

Move the fraction on the RHS to the LHS.
\[\frac{5}{x^2-3 x+2}-\frac{1}{x-2}-\frac{1}{3 x-3}=0 \]and then combine fractions.\[-\frac{4 (x-5)}{3 (x-2) (x-1)}=0 \]You should conclude that x = 5 from the resulting numerator.
A plot is attached.

Answer 9

So for an actual step by step solution:

0. Original Problem: \[\frac{5}{x^2 - 3x + 2} - \frac{1}{x-2} = \frac{1}{3x - 3}\]1. Express the fractions in factored form: \[\frac{5}{(x-2)(x-1)} - \frac{1}{x-2}=\frac{1}{3(x-1)}\]2. Add 1/(x+2) to both sides: \[\frac{5}{(x-2)(x-1)} = \frac{1}{3(x-1)} + \frac{1}{x-2}\]3. Split the fraction on the LHS to a product of 2 fractions: \[\frac{5}{x-2}\dot\ \frac{1}{x-1} = \frac{1}{3(x-1)}+ \frac{1}{x-2}\]4. Multiply both sides by (x-1): \[\frac{5}{x-2} = \frac{1}{3} + \frac{x-1}{x-2}\]5. Subtract (x-1)/(x-2) from both sides:\[\frac{5}{x-2}- \frac{x-1}{x-2} = \frac{1}{3}\]6. Combine the LHS: \[\frac{5-x+1}{x-2}=\frac{1}{3}\]7. Simplify the LHS: \[\frac{6-x}{x-2}=\frac{1}{3}\]8. Cross Multiply: \[3(6-x) = x-2\]9. Distribute: \[18 - 3x = x-2\]10. Place like terms on the same side: \[18 + 2 = x + 3x\]11. Simplify: \[20 = 4x\]12. Divide both sides by 4: \[\frac{20}{4} = x\]13. Solve for x: \[5 = x\]

Answer 10

wow thanks so much! that helped tremendously

Answer 11

Awesome =]

Answer 12

do you think you can help me with this one and also provide an explanation? @Hero

Answer 13

I'm about to go take a nap, but you can re-post this as a separate question and others should surely be capable of helping you.

Answer 14

okay, thanks so much though. you helped a lot! (: