1. ## Help with equations.

$\frac {4x^2 - x + 1}{2x^2 + 5x-3} - \frac{2x-9}{x+3} = \frac{5x-4}{2x-1} + \frac {7}{4}$

and this one(sorry i dont know how to put into code)
Fourth root [ 2 squareroot(2x-1) + squareroot(x+3) ]= Fourth root of [2 squareroot(3x+1)]

Hope you guys got that. Thanks!

2. Since you are unable to make any start, it would appear that you need first to learn about these equation types. For the first exercise, try here. For the second one, try here.

Once you have learned the basics of how to solve rational and radical equations, including learning how to solve quadratic equations, if necessary, please attempt the exercises. If you get stuck, you can then reply with a clear listing of your efforts so far.

Once we can "see" where you're having trouble, we can try to help you get "un-stuck".

3. $2x^2+5x-3 = (x+3)(2x-1)$

$
\frac {4x^2 - x + 1}{2x^2 + 5x-3} - \frac{2x-9}{x+3} = \frac{5x-4}{2x-1} + \frac {7}{4}
$

$\frac{4x^2-x+1}{(x+3)(2x-1)} - (\frac{2x-9}{x+3} \times \frac{2x-1}{2x-1}) = \frac{5x-4}{2x-1} \times \frac{x+3}{x+3} + \frac{7}{4} \times \frac{(x+3)(2x-1)}{(x+3)(2x-1)}$

Combining:

$\frac{4x^2-x+1 - (4x^2-2x-18x+9)}{(x+3)(2x-1)} = \frac{5x^2+15x-4x-12+1.75(2x^2+5x-3)}{(x+3)(2x-1)}$

The denominators can now be cancelled. Don't forget that subtracting a negative is the same as adding a positive: $a-(-b) = a+b$

4. Originally Posted by e^(i*pi)
$2x^2+5x-3 = (x+3)(2x-1)$

$
\frac {4x^2 - x + 1}{2x^2 + 5x-3} - \frac{2x-9}{x+3} = \frac{5x-4}{2x-1} + \frac {7}{4}
$

$\frac{4x^2-x+1}{(x+3)(2x-1)} - (\frac{2x-9}{x+3} \times \frac{2x-1}{2x-1}) = \frac{5x-4}{2x-1} \times \frac{x+3}{x+3} + \frac{7}{4} \times \frac{(x+3)(2x-1)}{(x+3)(2x-1)}$

Combining:

$\frac{4x^2-x+1 - (4x^2-2x-18x+9)}{(x+3)(2x-1)} = \frac{5x^2+15x-4x-12+1.75(2x^2+5x-3)}{(x+3)(2x-1)}$

The denominators can now be cancelled. Don't forget that subtracting a negative is the same as adding a positive: $a-(-b) = a+b$
Hey, isn't it that the LCD should be (4)(2x-1)(x+3)

Edit: Oh, you used 7/4 as 1.75

I think it's harder to use 1.75 so I also removed the 4 as a denominator.

Then I got this:
$34x^2+3x-37$

so from there, I should use the quadratic formula to solve for x, am i right?