# [SOLVED] HW3-QST1B- Albegra equation &amp; Validation -&gt; Please check if right?

• Aug 23rd 2008, 10:48 AM
Neenoon
[SOLVED] HW3-QST1B- Albegra equation &amp; Validation -&gt; Please check if right?
Here is one more that is not right I think but I can't figure anything else? (Angry)

$\frac {x+3}{2x-7}-\frac {2x-1}{x-3}=0$

I have to solve this and show validation.

Here is what I did (or tried to do):

$\frac {(x+3)}{(2x-7)}=\frac {(x+3)}{(2x+1)}$

$(x+3)(2x+1)=(2x-7)(x+3)$

$2x^2+x+6x+3=2x^2+6x-7x-21$

$2x^2+x+6x-2x^2-6x+7x=-21-3$

$8x=-24$

$x=-3$

Validation:

$\frac {-3+3}{2(-3)-7} = \frac {-3+3}{2(-3)+1}$

$\frac {0}{-13}=\frac {0}{-5}$

$0=0$
• Aug 23rd 2008, 11:02 AM
Chris L T521
Quote:

Originally Posted by Neenoon
Here is one more that is not right I think but I can't figure anything else? (Angry)

$\frac {x+3}{2x-7}-\frac {2x-1}{x-3}=0$

I have to solve this and show validation.

Here is what I did (or tried to do):

$\color{red}\frac {(x+3)}{(2x-7)}=\frac {(x+3)}{(2x+1)}$ This is where you went wrong.

$(x+3)(2x+1)=(2x-7)(x+3)$

$2x^2+x+6x+3=2x^2+6x-7x-21$

$2x^2+x+6x-2x^2-6x+7x=-21-3$

$8x=-24$

$x=-3$

Validation:

$\frac {-3+3}{2(-3)-7} = \frac {-3+3}{2(-3)+1}$

$\frac {0}{-13}=\frac {0}{-5}$

$0=0$

Your equation should have been $\frac {x+3}{2x-7}=\frac {2x-1}{x-3}$.

We then see that $(x+3)(x-3)=(2x-1)(2x-7)$

Try to solve the equation from here. Let us know what you get for $x$.

--Chris
• Aug 24th 2008, 02:38 AM
Neenoon
Quote:

Originally Posted by Chris L T521
Your equation should have been $\frac {x+3}{2x-7}=\frac {2x-1}{x-3}$.

We then see that $(x+3)(x-3)=(2x-1)(2x-7)$

Try to solve the equation from here. Let us know what you get for $x$.

--Chris

That's what I tried to do first, but I can't solve it like that... let's see...

$(x+3)(x-3)=(2x-1)(2x-7)$

$x^2-3x+3x-9=4x^2-14x-2x+7$

$x^2-4x^2+16x=9+7$

$-3x^2+16x=16$

And then I'm stuck with that $-3x^2$ ???

What am I not getting here?
• Aug 24th 2008, 02:45 AM
Spec
Factor out 3 and the equation becomes:

$x^2 - \frac{16x}{3} + \frac{16}{3} = 0$
• Aug 24th 2008, 03:08 AM
Neenoon
I'm sorry but I'm still not getting how am I suppose to get the value of x from that?

Quote:

Originally Posted by Spec
Factor out 3 and the equation becomes:

$x^2 - \frac{16x}{3} + \frac{16}{3} = 0$

• Aug 24th 2008, 03:26 AM
Spec
Haven't you learned how to solve quadratic equations yet? It's not that hard really.

Quadratic equation - Wikipedia, the free encyclopedia
• Aug 24th 2008, 08:36 AM
Neenoon
Here is what I did, tell me if this is right?

$(x+3)(x-3)=(2x-1)(2x-7)$

$x^2-3x+3x-9=4x^2-2x-14x+7$

$x^2-4x^2-3x+3x+2x+14x-9-7=0$

$-3x^2+16x-16=0$

So, a=-3 b=16 c=-16
$
x=-b \pm \frac{\sqrt{b^2-4ac}}{2a}$

$
x=-16 \pm \frac{\sqrt{16^2-(4 \bullet -3 \bullet 16)}}{2(-3)}$

$
x=-16 \pm \frac{\sqrt{64}}{-6)}$

$
x=\frac{-16+8}{-6}$

and
$
x=\frac{-16-8}{-6}$

$x=\frac{4}{3}\ and\ x=4$
• Aug 24th 2008, 09:02 AM
Chop Suey
Brave! You're correct.

But you should write the -b in the numerator not outside.
• Aug 26th 2008, 10:20 AM
Neenoon
Thank you, will do!