Algebraic Fractions: Is this method correct?

**vbiter** · October 23rd 2009, 08:41 AM

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

$Solve for x$

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

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

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

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

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

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

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

**Defunkt** · October 23rd 2009, 08:50 AM

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

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

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

**vbiter** · October 23rd 2009, 04:17 PM

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:

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

and could you clarify this?

Thanks

**user_5** · October 23rd 2009, 06:41 PM

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

Cross multiplying gives you:

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

Expanding the LHS and collecting like terms

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

Use the quadratic formula to find the solutions.

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

**HallsofIvy** · October 24th 2009, 06:06 AM

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

$Solve for x$

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

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

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

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 $8/5 = 2x-3 => 1$ $3/5 = 2x-3$

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

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

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

**Defunkt** · October 24th 2009, 08:27 AM

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:

$ \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.

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