Subtract: (b^2 - 6a^2) / (a - b) - (5a^2) / (b - a)

Re: Subtract

First you need to learn to use parenthesis to say what you mean.

Your expression ( b^2-6a^2 / a-b ) - ( 5a^2 / b-a ) means \(\displaystyle (b^2-\frac{6a^2}{a}-b) - (\frac{5a^2}{b}-a)\).

I'll assume you mean ( b^2-6a^2 / (a-b) ) - ( 5a^2 / (b-a) )


If that's the case, change -5a^2 /( b-a) to +5a^2 /( a-b)
 
Re: Subtract

You are correct that is what I meant. Sorry. I think I have it now, looks like -a-b. Thanks
 
Re: Subtract

algebraiseasy said:
You are correct that is what I meant. Sorry. I think I have it now, looks like -a-b. Thanks
Click to expand...

It's NOT -a-b.
 
algebraiseasy said:
You are correct that is what I meant. Sorry. I think I have it now, looks like -a-b. Thanks
Click to expand...
Try using the solution set-up provided by the previous reply:

Since b - a = -(a - b), and since the other denominator is already (a - b), try reversing the subtraction in the "(b - a)" denominator, multiplying the sign change (the -1) through the numerator, and then combining the two fractions the tutor gave you. :wink:
 
I have the numerator as b^2 - 6a^2 + 5a^2 and the denominator as a-b after canceling I get -a-b. No?
 
algebraiseasy said:
( b^2-6a^2 / a-b ) - ( 5a^2 / b-a )
What is my LCD?
Would it be ( a-b )( b-a )?
Click to expand...

Enough time has passed and sufficient work has been shown - so here is the detailed answer....

I think your problem is:

\(\displaystyle \frac{b^2 - 6a^2}{a-b} \, - \, \frac{5a^2}{b-a}\)

\(\displaystyle = \, \frac{b^2 - 6a^2}{a-b} \, + \, \frac{5a^2}{a-b}\)

\(\displaystyle = \, \frac{(b^2 - 6a^2) \, + \, 5a^2}{a-b}\)

\(\displaystyle = \, \frac{b^2 - a^2}{a-b}\)

\(\displaystyle = \,- \, \frac{(a -b)\cdot (a+b)}{a-b}\)

\(\displaystyle = \,- \, (a+b)\)

\(\displaystyle = \,- a \, - \, b\)
 
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