Subtract and Simplify

[NoGoodAtMath]

So here is the problem. (3x^2+2x/x-1) - (10x-5/x-1)

I know they both have the the same denominator, so do I just simply subtract what's on top?

 

[JeffM]

So here is the problem. (3x^2+2x/x-1) - (10x-5/x-1)

I know they both have the the same denominator, so do I just simply subtract what's on top?

In the future, please be careful about parentheses. I suspect you meant: (3x^2 + 2x)/(x - 1) - (10x - 5)/(x - 1).

The thing to remember about elementary algebra is that the letters and expressions just stand for ordinary numbers. So everything you learned in arithmetic applies in elementary algebra.

$\displaystyle \dfrac{8}{11} - \dfrac{5}{11} = \dfrac{3}{11}.$

The rule you learned in arithmetic about subtracting two fractions with a COMMON denominator was: $\displaystyle \dfrac{a}{b} - \dfrac{c}{b} = \dfrac{a - c}{b}.$

So, yes, in the problem that you asked about, you simply subtract the numerators because the denominator of both fractions is the same.

$\displaystyle \dfrac{3x^2 + 2x}{x - 1} - \dfrac{10x - 5}{x - 1} = \dfrac{3x^2 - 8x + 5}{x - 1}.$

Question: can you simplify that expression?

 

[NoGoodAtMath]

In the future, please be careful about parentheses. I suspect you meant: (3x^2 + 2x)/(x - 1) - (10x - 5)/(x - 1).

The thing to remember about elementary algebra is that the letters and expressions just stand for ordinary numbers. So everything you learned in arithmetic applies in elementary algebra.

$\displaystyle \dfrac{8}{11} - \dfrac{5}{11} = \dfrac{3}{11}.$

The rule you learned in arithmetic about subtracting two fractions with a COMMON denominator was: $\displaystyle \dfrac{a}{b} - \dfrac{c}{b} = \dfrac{a - c}{b}.$

So, yes, in the problem that you asked about, you simply subtract the numerators because the denominator of both fractions is the same.

$\displaystyle \dfrac{3x^2 + 2x}{x - 1} - \dfrac{10x - 5}{x - 1} = \dfrac{3x^2 - 8x + 5}{x - 1}.$

Question: can you simplify that expression?

I think I should use the AC Method and then group. So what I was left with was an (x+1)(3x-5) Is that correct?

 

[Deleted member 4993]

I think I should use the AC Method and then group. So what I was left with was an (x+1)(3x-5) Is that correct?

No your answer should be:

$\displaystyle \dfrac{(x-1)(3x-5)}{(x-1)} \ = \ 3x - 5$