# Thread: Algebraic Fractions: Is this method correct?

1. ## Algebraic Fractions: Is this method correct?

I am having trouble finding the correct method for solving algebraic fractions. I have tried the sequence of operations below and it appears to fit, but would appreciate some help / guidance on a robust process for working through to solve these equations. Thanks

$\displaystyle Solve for x$

$\displaystyle x+5/x - 8/2x-3 = 0$

can be rewritten as $\displaystyle x+5/x = 8/2x-3$

which cancels out to $\displaystyle 5 = 8/2x-3$

which can be rewritten as $\displaystyle 8/5 = 2x-3 => 1$ $\displaystyle 3/5 = 2x-3$

and therefore $\displaystyle 2x = 1 3/5 + 3$

and $\displaystyle x = 1/2(4$ $\displaystyle 3/5)$

==> $\displaystyle x = 2$ $\displaystyle 3/10$

2. It would really help if you wrote the fractions in a non-ambiguous way... did you mean $\displaystyle \frac{x+5}{x} -\frac{8}{2x-3} = 0$? or $\displaystyle x + \frac{5}{x} - \frac{8}{2x} -3 = 0$? or something else? It is very hard to understand...

If you meant the first, then: $\displaystyle \frac{x+5}{x} = \frac{8}{2x-3} \Rightarrow (x+5)(2x-3) = 8x \Rightarrow 2x^2 +7x -15 = 8x \Rightarrow 2x^2 -x -15 = 0$

Now solve for x and that is your solution (use the quadratics formula $\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$)

3. Thank you for replying, (you read the equation correctly) but your response is a little unclear beyond the middle section.

Cross multiplying to obtain (x+5)(2x-3) = 8x makes perfect sense, as does expanding the brackets on the left side. However I can't follow how 8x translates to 16x - 24

I appears as if the 8x value is discarded, and 8 is re multiplied against 2x-3

Would this be what it is supposed to look like:

$\displaystyle \frac{(x+5)(2x-3)}{x} = \frac{8(2x-3)}{2x-3}$

and could you clarify this?

Thanks

4. $\displaystyle \frac{x+5}{x} =\frac{8}{2x-3}$

Cross multiplying gives you:

$\displaystyle (x+5)(2x-3) = 8x$

Expanding the LHS and collecting like terms

$\displaystyle 2x^2-3x+10x-15= 8x \Rightarrow 2x^2-x-15=0$

Use the quadratic formula to find the solutions.

Which are $\displaystyle x= -\frac{5}{2}$ or $\displaystyle x=3$

5. Originally Posted by vbiter
I am having trouble finding the correct method for solving algebraic fractions. I have tried the sequence of operations below and it appears to fit, but would appreciate some help / guidance on a robust process for working through to solve these equations. Thanks

$\displaystyle Solve for x$

$\displaystyle x+5/x - 8/2x-3 = 0$

can be rewritten as $\displaystyle x+5/x = 8/2x-3$

which cancels out to $\displaystyle 5 = 8/2x-3$
No, it doesn't. If you mean (x+5)/x, that is equal to 1+ 5/x, not 5.

Assuming that you mean (x+5)/x= 8/(2x-3), I would recommend that you get rid of the fractions by multiplying both sides by the denominators x(2x-3)

Multiplying the left side by x(2x-3) cancels the "x"s and gives (x+5)(2x-3). Multiplying the right side by x(2x-3) cancels the "2x-3"s and gives 8(x). So your equation becomes (x+5)(2x-3)= 8x. That is a quadratic equation for x.

which can be rewritten as $\displaystyle 8/5 = 2x-3 => 1$ $\displaystyle 3/5 = 2x-3$

and therefore $\displaystyle 2x = 1 3/5 + 3$

and $\displaystyle x = 1/2(4$ $\displaystyle 3/5)$

==> $\displaystyle x = 2$ $\displaystyle 3/10$

6. Originally Posted by vbiter
Thank you for replying, (you read the equation correctly) but your response is a little unclear beyond the middle section.

Cross multiplying to obtain (x+5)(2x-3) = 8x makes perfect sense, as does expanding the brackets on the left side. However I can't follow how 8x translates to 16x - 24

I appears as if the 8x value is discarded, and 8 is re multiplied against 2x-3

Would this be what it is supposed to look like:

$\displaystyle \frac{(x+5)(2x-3)}{x} = \frac{8(2x-3)}{2x-3}$

and could you clarify this?

Thanks
Oops, don't know what happened there... fixed the original post now.