# Equivalent Fractions [Solved!]

**Taradawn** 15 Dec 2015, 03:04

### My question

HELP 6x^2+13x-5/6x^3-2x^2

divide (3x-1) to denominator and numerator

please show me how to solve this

### Relevant page

5. Equivalent Fractions

### What I've done so far

Tried to do it sseveral ways but got stuck

X

HELP 6x^2+13x-5/6x^3-2x^2
divide (3x-1) to denominator and numerator
please show me how to solve this

Relevant page
<a href="/factoring-fractions/5-equivalent-fractions.php">5. Equivalent Fractions</a>
What I've done so far
Tried to do it sseveral ways but got stuck

## Re: Equivalent Fractions

**Newton** 15 Dec 2015, 13:25

Hello Taradawn

You can factor both the top and the bottom of that fraction. (A big hint is that (3x-1) will go into both top and bottom.)

Your problem is a bit like the one at the very bottom of this page:

6. Multiplication and Division of Fractions

You factor everything first, then you can divide top and bottom.

Can you go from there?

BTW, if you use the math input system, you can make much more readable math, like this:

`\frac{6x^2+13x-5}{6x^3-2x^2}`

Regards

X

Hello Taradawn
You can factor both the top and the bottom of that fraction. (A big hint is that (3x-1) will go into both top and bottom.)
Your problem is a bit like the one at the very bottom of this page:
<a href="/factoring-fractions/6-multiplication-division-fractions.php">6. Multiplication and Division of Fractions</a>
You factor everything first, then you can divide top and bottom.
Can you go from there?
BTW, if you use the math input system, you can make much more readable math, like this:
`\frac{6x^2+13x-5}{6x^3-2x^2}`
Regards

## Re: Equivalent Fractions

**Taradawn** 15 Dec 2015, 22:00

i try

`6x^2+13x-5/6x^3-2x^2` ` = (3x - 1)(2x + 5)/2x^2(3x-1)`

But when I do "Preview", the math looks bad. Why?

X

i try
`6x^2+13x-5/6x^3-2x^2` ` = (3x - 1)(2x + 5)/2x^2(3x-1)`
But when I do "Preview", the math looks bad. Why?

## Re: Equivalent Fractions

**Murray** 16 Dec 2015, 16:27

Your answer so far is good, but you need to remember to use brackets so the fractions work properly.

Instead of

`6x^2+13x-5/6x^3-2x^2 = (3x - 1)(2x + 5)/2x^2(3x-1)`

it should be

`(6x^2+13x-5)/(6x^3-2x^2) = ((3x - 1)(2x + 5))/(2x^2(3x-1))`

so it looks like

`(6x^2+13x-5)/(6x^3-2x^2) = ((3x - 1)(2x + 5))/(2x^2(3x-1))`

X

Your answer so far is good, but you need to remember to use brackets so the fractions work properly.
Instead of
<code>6x^2+13x-5/6x^3-2x^2 = (3x - 1)(2x + 5)/2x^2(3x-1)</code>
it should be
<code>(6x^2+13x-5)/(6x^3-2x^2) = ((3x - 1)(2x + 5))/(2x^2(3x-1))</code>
so it looks like
`(6x^2+13x-5)/(6x^3-2x^2) = ((3x - 1)(2x + 5))/(2x^2(3x-1))`

## Re: Equivalent Fractions

**Taradawn** 17 Dec 2015, 15:10

OK, got it.

I've factored, so now I can cancel:

`(6x^2+13x-5)/(6x^3-2x^2)` ` = ((3x - 1)(2x + 5) -: (3x-1))/(2x^2(3x-1) -: (3x-1))` ` = (2x+5)/(2x^2)`

Am I right?

X

OK, got it.
I've factored, so now I can cancel:
`(6x^2+13x-5)/(6x^3-2x^2)` ` = ((3x - 1)(2x + 5) -: (3x-1))/(2x^2(3x-1) -: (3x-1))` ` = (2x+5)/(2x^2)`
Am I right?

## Re: Equivalent Fractions

**Murray** 18 Dec 2015, 07:07

Yes, well done!

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