Equivalent Fractions [Solved!]

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

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

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?

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))`

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?

X

Yes, well done!