# proof - binomial fraction simplifies to a constant

• Nov 6th 2012, 02:44 PM
indymogul
proof - binomial fraction simplifies to a constant
Hi, just starting chapter review of Algebra 1. The first question is:

Show that the following simplifies to a constant when x =/= 1/2:

x - 1 3 - 7x
------- + -------
2x - 1 1 - 2x

I have no idea how to tackle this. I can see that if x = 1/2, you're going to be dividing by zero on both sides, which will give non-real numbers (imaginary?) because anything divided by zero is like, infinity.

BUT how do you show it simplifies to a constant? Do I just find a common denominator, add and simplify? Is that it?

• Nov 6th 2012, 03:29 PM
Prove It
Re: proof - binomial fraction simplifies to a constant
Quote:

Originally Posted by indymogul
Hi, just starting chapter review of Algebra 1. The first question is:

Show that the following simplifies to a constant when x =/= 1/2:

x - 1 3 - 7x
------- + -------
2x - 1 1 - 2x

I have no idea how to tackle this. I can see that if x = 1/2, you're going to be dividing by zero on both sides, which will give non-real numbers (imaginary?) because anything divided by zero is like, infinity.

BUT how do you show it simplifies to a constant? Do I just find a common denominator, add and simplify? Is that it?

This is nearly impossible to read. Is it \displaystyle \displaystyle \begin{align*} \frac{x - 1}{2x - 1} + \frac{3 - 7x}{1 - 2x} \end{align*}?
• Nov 6th 2012, 11:05 PM
indymogul
Re: proof - binomial fraction simplifies to a constant
Sorry about that. some of the spacing was removed from the post.

Yes, you got it spot on. It's exactly how you have it written. How did you do that, by the way? I'm having trouble formatting equations etc. correctly.

• Nov 7th 2012, 12:20 AM
cac2008
Re: proof - binomial fraction simplifies to a constant
$\displaystyle \frac{x-1}{2x-1}+\frac{3-7x}{1-2x}$

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

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

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

$\displaystyle x-1=3-7x$

$\displaystyle 8x=4$

$\displaystyle x=2$
• Nov 7th 2012, 12:30 AM
cac2008
Re: proof - binomial fraction simplifies to a constant
And with regards to the formatting, do a google search for LATEX. It seems like a minefield to cross, but for help with the simple stuff just hover your mouse over some equations on this forum and you should get a tooltip appearing showing the formatting that made the displayed equation.
• Nov 7th 2012, 01:50 AM
indymogul
Re: proof - binomial fraction simplifies to a constant
Thanks. Can you explain the steps here? I don't get why you equate the two fractions where the one on the right becomes negative, and then why you switch the order on the bottom with the RHS fraction, etc. I don't really understand what's happening. Could you give me a brief note for each step? Also how this shows that it simplifies to a constant? If I understand the theory I will be able to solve problems like this without help. Thank you for your help.
• Nov 7th 2012, 01:58 AM
MarkFL
Re: proof - binomial fraction simplifies to a constant
$\displaystyle \frac{x-1}{2x-1}+\frac{3-7x}{1-2x}=\frac{x-1}{2x-1}+\frac{7x-3}{2x-1}=\frac{8x-4}{2x-1}=\frac{4(2x-1)}{2x-1}=4$
• Nov 7th 2012, 02:01 AM
indymogul
Re: proof - binomial fraction simplifies to a constant
Now I get it. So you basically simplify and collect like terms. It simplifies to a constant because 4 over 1 is left when you remove the common factor from top and bottom. Is that right? Thank you very much!
• Nov 7th 2012, 02:24 AM
MarkFL
Re: proof - binomial fraction simplifies to a constant
Yes, that's right.