1. ## Algebraic fractions

x^2 - 25/25 + 5x

I can factorize the numerator but not the denominator
I'm familiar with factorizing, canceling and expanding but I've never confronted something like 5x + 25 to expand/factorize.

2. ## Re: Algebraic fractions

(2.5 x 10^9)/(5 x 10^3)

in standard form. this isn't algebra but I'm hoping anybody has the patience to answer both.

Write as a single fraction in it's simplest form:

2x/x-1 - 7x-3/x^2-1

Please explain how you got these answers too, in the most efficient method.

Thanks!

3. ## Re: Algebraic fractions

Originally Posted by Ashir

x^2 - 25/25 + 5x

I can factorize the numerator but not the denominator
I'm familiar with factorizing, canceling and expanding but I've never confronted something like 5x + 25 to expand/factorize.
$5x+25$, these two terms have a common factor $5$ thus $5x+25 = 5(x+5)$

4. ## Re: Algebraic fractions

Originally Posted by Ashir
(2.5 x 10^9)/(5 x 10^3)
2.5 * 10^9 / (5 * 10^3)
= 2.5 * 10^9 * 10^-3 / 5
= 5/2 * 10^6 / 5
= 5 * 10^6 / 10
= 5 * 10^5

Hope we're not doing your homework!!

5. ## Re: Algebraic fractions

Originally Posted by Ashir

x^2 - 25/25 + 5x

I can factorize the numerator but not the denominator
I'm familiar with factorizing, canceling and expanding but I've never confronted something like 5x + 25 to expand/factorize.
Please use brackets where they're needed, or else learn some LaTeX. As written, your expression is \displaystyle \begin{align*} x^2 - \frac{25}{25} + 5x \end{align*}...

6. ## Re: Algebraic fractions

Originally Posted by Ashir
...

Write as a single fraction in it's simplest form:

2x/x-1 - 7x-3/x^2-1

I am assuming you mean:

$\frac{2x}{x-1}-\frac{7x-3}{x^2-1}$

First, factor the denominators:

$\frac{2x}{x-1}-\frac{7x-3}{(x+1)(x-1)}$

Now it is clear what we need to multiply the numerator and denominator of the first term with to get a common denominator:

$\frac{2x}{x-1}\cdot\frac{x+1}{x+1}-\frac{7x-3}{(x+1)(x-1)}$

Can you finish?

7. ## Re: Algebraic fractions

Originally Posted by MarkFL2
Can you finish?
No, I'm too busy. Could you finish it for me?

8. ## Re: Algebraic fractions

Originally Posted by Wilmer
No, I'm too busy. Could you finish it for me?
Ding! Fries are done.

9. ## Re: Algebraic fractions

Originally Posted by Siron
$5x+25$, these two terms have a common factor $5$ thus $5x+25 = 5(x+5)$
How could I not have saw that? Thanks.

Originally Posted by Wilmer
2.5 * 10^9 / (5 * 10^3)
= 2.5 * 10^9 * 10^-3 / 5
= 5/2 * 10^6 / 5
= 5 * 10^6 / 10
= 5 * 10^5

Hope we're not doing your homework!!
Not following you. This area is very new to me, in fact we just started this week.

Nope, you're correcting the questions I didn't get right in an exam :0)

Originally Posted by MarkFL2
I am assuming you mean:

$\frac{2x}{x-1}-\frac{7x-3}{x^2-1}$

First, factor the denominators:

$\frac{2x}{x-1}-\frac{7x-3}{(x+1)(x-1)}$

Now it is clear what we need to multiply the numerator and denominator of the first term with to get a common denominator:

$\frac{2x}{x-1}\cdot\frac{x+1}{x+1}-\frac{7x-3}{(x+1)(x-1)}$

Can you finish?
2X x X+1 = 2X+2 right? Or does the constant multiplied with the coefficient of the first term only apply to brackets?

And otherwise I'm lost on what I do next. We're revisiting this area after a year or so.

10. ## Re: Algebraic fractions

Originally Posted by Ashir
2X x X+1 = 2X+2 right?
Please use * for multiplication sign; and small "x" for variable...; and brackets when necessary...

2x * (x + 1) = 2x^2 + 2x ; everything inside brackets is multiplied by what's outside...

11. ## Re: Algebraic fractions

Forgot to times 2x by x sorry. Lost a good few marks on the exam because of clumsy mistakes

12. ## Re: Algebraic fractions

Originally Posted by Ashir
Forgot to times....
to MULTIPLY....

13. ## Re: Algebraic fractions

We say times here in England :P

14. ## Re: Algebraic fractions

Originally Posted by cobooboc
I do not know why I proposed to be deleted, but I really do not understand that, can anyone give me an explanation?Or we do not have a common language.

15. ## Re: Algebraic fractions

Back on topic.

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