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# 1 variable equation

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Find all real values of $x$ such that $$\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}.$$Enter all the solutions, separated by commas.

Dec 26, 2020

#1
+1

(2x^2 - 3x)/(x^2 - x) + 5x - 11 = (3x^2 + 5x + 2)/(x^2 - 1)

==> (2x^2 - 3x)(x^2 - 1) + (5x - 11)(x^2 - x)(x^2 - 1) = (3x^2 + 5x + 2)(x^2 - x)

==> 5x^2 - 19x^4 + 7x^3 + 19x^2 - 12x = 0

==> x(x - 1)(x + 1)(5x - 4)(x - 3) = 0

Solutions are 0, 1, -1, 4/5, 3

Dec 26, 2020
#2
-2

This must be incorrect, since if you plug in 1 and 0, the denominator will be zero resulting in undefined. Wrong

Dec 26, 2020
#4
+112000
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Did you bother to check if the other answers are correct?

Or are you just here to criticize?

Melody  Dec 26, 2020
#3
+112000
+2

$$\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\ \frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\ x\ne\pm1,\quad x\ne0\\ [\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\ \frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\ (2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\ (2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\ (2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\ 5x^3 -9x^2-6x+8= 3x^2+5x+2 \\ 5x^3 -12x^2-11x+6= 0 \\ (x-3)(5x-2)(x+1)=0\\ \text{x can't equal -1 so}\\ x=3 \;\;or\;\;x=0.4$$

LaTex

\frac{2x^2-3x}{x^2-x} + 5x -11 = \frac{3x^2+5x+2}{x^2-1}\\
\frac{2x^2-3x}{x(x-1)} + 5x -11 = \frac{3x^2+5x+2}{(x-1)(x+1)}\\
[\frac{2x^2-3x}{x(x-1)} + 5x -11]*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{(x-1)(x+1)}]*[x(x-1)(x+1)] \\
\frac{2x^2-3x}{x(x-1)} *[x(x-1)(x+1)]+ (5x -11)*[x(x-1)(x+1)] = [\frac{3x^2+5x+2}{1}]*[x] \\
(2x^2-3x)(x+1)+ (5x -11)(x(x^2-1))=x (3x^2+5x+2) \\
(2x-3)(x+1)+ (5x -11)(x^2-1)= (3x^2+5x+2) \\
(2x^2-x-3)+ (5x^3 -11x^2-5x+11)= (3x^2+5x+2) \\
5x^3 -9x^2-6x+8= 3x^2+5x+2 \\
5x^3 -12x^2-11x+6= 0 \\
(x-3)(5x-2)(x+1)=0\\
\text{x can't equal -1  so}\\
x=3 \;\;or\;\;x=0.4

Dec 26, 2020
edited by Melody  Dec 26, 2020
edited by Melody  Dec 27, 2020
#5
0

Yes, I checked, since seeing the first three solutions I immediately saw and noticed, the strategy was alright, but the person needed to recheck and see if it was valid, I am not here to criticize. When I do my work, I always check.

Dec 27, 2020
#6
+112000
0

Yes the first three answers should have been discarded but did you bother to check the next 2 answers.

You still have not said that you have?  Are they right?  Are they wrong?

If you knew some things were correct then you should have said so.

You can politely suggest to people that maybe they should check their answer by substitution without being rude about it.

Melody  Dec 27, 2020
edited by Melody  Dec 27, 2020
#8
+112000
0

The respondent does not need to check their answer if they do not want to.

It is up to the asker to accept, reject or alter an answer as they feel the need.

The asker should at the very least check answers by substitution if that is possible.

The respondents are here to offer their help or suggestions.

It is not the respondent's job to give polished answers that can be copied without thought.

Melody  Dec 27, 2020
#7
0

👌

.
Dec 27, 2020
#9
0

Thank you for the suggestions, I will follow your rules and try my best to "respect", by the way, My results were that only one answer the guest provided was correct.

Dec 27, 2020
#10
+112000
0

Which one.  I am sure the answerer would like some positive feedback from you.

Melody  Dec 27, 2020
#11
+112000
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Yes you are right.

One of the answers is correct.

Now the answerer has given you a technique that you say you understand.

And he/she has given you one correct answer.

That correct answer makes it easier for you to solve it by yourself.  (I know because i have solve it)

So you have a technique and a correct answer.

Seems to me that the guest answerer has helped you are real lot!

Melody  Dec 27, 2020
#12
0

Yes, 3 is the only correct answer, I found only 3 and 2/5 are the correct answer!

Dec 27, 2020
#13
+112000
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Yes that is what I got too.  I might show my answer now.