Help on Algebraic fractions? Thank you :]

[Lexadis]

I have done the questions, but I wanted to know whether they are correct, since I'm sort of weak when it comes to algebraic fractions
I have two questions, and it would be a very great help if you guys can check it out, and thank you very very much in advance (and hope I'm not much of a both ^^)
Q1.
$\displaystyle \frac{6x}{x^{2}-4x+3}+\frac{2}{4x-12} = \frac{6x}{\left ( x-1 \right )\left ( x-3 \right )}+\frac{2}{4\left ( x-3 \right )}$

.......................$\displaystyle =\frac{4\left ( 6x \right )+2\left ( x-1 \right )}{4\left ( x-3 \right )\left ( x-1 \right )}$

.......................$\displaystyle =\frac{24x+2x-2}{4\left ( x-3 \right )\left ( x-1 \right )}$

.......................$\displaystyle =\frac{26x-2}{4\left ( x-3 \right )\left ( x-1 \right )}$

.......................$\displaystyle =\frac{2\left ( 13x-1 \right )}{4\left ( x-3 \right )\left ( x-1 \right )}$

.......................$\displaystyle =\frac{13x-1}{2\left ( x-3 \right )\left ( x-1 \right )}$

Correct?

And Q2:

$\displaystyle \frac{a-3}{a^{2}-3a-4}-\frac{a-1}{a^{2}-a-2} = \frac{a-3}{\left ( a+1 \right )\left ( a-4 \right )}-\frac{a-1}{\left ( a+1 \right )\left ( a-2 \right )}$

.......................$\displaystyle =\frac{\left ( a-3 \right )\left ( a-2 \right )-\left ( a-1 \right )\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{a\left ( a-2 \right )-3\left ( a-2 \right )-a\left ( a-4 \right )-1\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{a^{2}-2a-3a+6-a^{2}+4a-a+4}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{10-2a}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{2\left ( 5-a \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

?

 

[DrPhil]

I have done the questions, but I wanted to know whether they are correct, since I'm sort of weak when it comes to algebraic fractions
I have two questions, and it would be a very great help if you guys can check it out, and thank you very very much in advance (and hope I'm not much of a both ^^)
Q1............$\displaystyle =\frac{13x-1}{2\left ( x-3 \right )\left ( x-1 \right )}$

Correct? YES

And Q2:

$\displaystyle \frac{a-3}{a^{2}-3a-4}-\frac{a-1}{a^{2}-a-2} = \frac{a-3}{\left ( a+1 \right )\left ( a-4 \right )}-\frac{a-1}{\left ( a+1 \right )\left ( a-2 \right )}$

.......................$\displaystyle =\frac{\left ( a-3 \right )\left ( a-2 \right )-\left ( a-1 \right )\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{a\left ( a-2 \right )-3\left ( a-2 \right )-a\left ( a-4 \right )-1\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$......OOPS
?

Sign error .. final term in the numerator should be +1(a - 4).
The "a" terms will also cancel out, so the numerator will be a constant (not 10, though).

 

[Lexadis]

Sign error .. final term in the numerator should be +1(a - 4).
The "a" terms will also cancel out, so the numerator will be a constant (not 10, though).

Ooh thank you c:
So...

$\displaystyle \frac{a-3}{a^{2}-3a-4}-\frac{a-1}{a^{2}-a-2} = \frac{a-3}{\left ( a+1 \right )\left ( a-4 \right )}-\frac{a-1}{\left ( a+1 \right )\left ( a-2 \right )}$

.......................$\displaystyle =\frac{\left ( a-3 \right )\left ( a-2 \right )-\left ( a-1 \right )\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{a\left ( a-2 \right )-3\left ( a-2 \right )-a\left ( a-4 \right )+1\left ( a-4 \right )}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{a^{2}-2a-3a+6-a^{2}+4a+a-4}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

.......................$\displaystyle =\frac{2}{\left ( a+1 \right )\left ( a-2 \right )\left ( a-4 \right )}$

?

 

[Deleted member 4993]

While checking answer with the method mentined above:

Do NOT assign value to 'x' such that the denominator goes to zero.

So for your problem #1, do not choose x = 1 or x = 3

for your problem #2, do not choose a = 11 or a = 2 or a = 4

All other numbers are fair game.

I suggest checking with at least 2 numbers.

 

[lookagain]

While checking answer with the method mentined above:

Do NOT assign value to 'x' such that the denominator goes to zero.

So for your problem #1, do not choose x = 1 or x = 3

for your problem #2, do not choose > > > a = 11 ** < < < or a = 2 or a = 4

All other numbers are fair game.

I suggest checking with at least 2 numbers.

**

$\displaystyle a \ = -1 \ \ was \ \ meant \ \ here.$


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You started with: $\displaystyle \frac{6x}{x^{2}-4x+3}+\frac{2}{4x-12} =$

and ended with: $\displaystyle =\frac{13x-1}{2\left ( x-3 \right )\left ( x-1 \right )}$
You can check YOURSELF if you're correct; simply assign a value to x
and substitute in both; as example if x = 5, then both will equal 4,
so you know you're correct...OK?

But suppose, for instance, that the final fraction for the alleged answer had come out to be:


$\displaystyle \ \dfrac{15x - 11}{2(x - 3)(x - 1)}.$


Although x = 5 makes that equal 4, it's not the correct fraction.

 

[Lexadis]

You started with: $\displaystyle \frac{6x}{x^{2}-4x+3}+\frac{2}{4x-12} =$

and ended with: $\displaystyle =\frac{13x-1}{2\left ( x-3 \right )\left ( x-1 \right )}$
You can check YOURSELF if you're correct; simply assign a value to x
and substitute in both; as example if x = 5, then both will equal 4,
so you know you're correct...OK?

Ooh okay didn't think of double checking that way :]

While checking answer with the method mentined above:

Do NOT assign value to 'x' such that the denominator goes to zero.

So for your problem #1, do not choose x = 1 or x = 3

for your problem #2, do not choose a = 11 or a = 2 or a = 4

All other numbers are fair game.

I suggest checking with at least 2 numbers.

Thank you. c:

**

$\displaystyle a \ = -1 \ \ was \ \ meant \ \ here.$


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



But suppose, for instance, that the final fraction for the alleged answer had come out to be:


$\displaystyle \ \dfrac{15x - 11}{2(x - 3)(x - 1)}.$


Although x = 5 makes that equal 4, it's not the correct fraction.

Got it