c-a^2/c/1-a/b

[chesterdg123]

This seems right?

b-a^2/c/1-a/b

=b.b/b -a^2/b/
1.b/b - a/b

=b/b-a^2/b/
b/b-a/b

=b-a^2/b/
b-a/b

=b(b-a)(b-a)/
b(b-a)

=b-a

 

[Denis]

This seems right?
b-a^2/c/1-a/b

=b.b/b -a^2/b/
1.b/b - a/b

=b/b-a^2/b/
b/b-a/b

=b-a^2/b/
b-a/b

=b(b-a)(b-a)/
b(b-a)

=b-a

1: what are you doing?!
2: where did the "c" go?
3: ask someone to show you how to post PROPERLY

 

[chesterdg123]

Sorry, this should be clearer.

Simplify


numerator b-a^2/b
denominator 1-a/b

b-a^2
1-a/b

=b(b/b)-a^2/b/
1(b/b) - a/b

=b/b-a^2/b
b/b-a/b

=b-a^2/b
b-a/b

=b(b-a)(b-a)
b(b-a)

=b-a

 

[stapel]

Sorry, this should be clearer.

numerator b-a^2/b
denominator 1-a/b

Is the numerator either of the following?

. . . . .b - (a^2)/b

. . . . .(b - a^2)/b

Or something else?

Is the denominator either of the following?

. . . . .1 - (a/b)

. . . . .(1 - a)/b

Or something else?

I'm afraid I can't make heads or tails of the rest of your post. Sorry.

Eliz.

 

[chesterdg123]

Sorry, it's :


b - (a^2/b)
1 - (a/b)

b-a^2
1-a/b

(=b(b/b)-(a^2/b)
1(b/b) - (a/b)

=(b/b)-(a^2/b)
(b/b)-(a/b)

=(b-a^2)/b
(b-a)/b

=b(b-a)(b-a)
b(b-a)

=b-a

Hopefully this is not too confusing

 

[stapel]

Sorry, it's :

b - (a^2/b)
1 - (a/b)

b-a^2
1-a/b

How did you go from:

. . . . .\(\displaystyle \frac{\left(b\, -\, \frac{a^2}{b}\right)}{\left(1\, -\, \frac{a}{b}\right)}\)

...to:

. . . . .\(\displaystyle \frac{\left(b\, -\, a^2\right)}{\left(1\, -\, \frac{a}{b}\right)}\)

Please clarify, using either LaTeX (as show above) or else the formatting explained in the articles at the links you saw in the "Read Before Posting" thread.

Thank you!

Eliz.

 

[chesterdg123]

Complex Rational Expressions

I'll try it again, please bear with me, I am new to writting equations this way.






b - (a^2/b)
----------
1 - (a/b)



b/1 . (b)/(b) - (a^2/b)
---------------------
1/1 . (b)/(b) -(a/b)


=( b^2-a^2)/b
--------------

(b-a)/b



=( b^2-a^2) /b (b/b-a)


= (b^3-a^2b)
b^2-ab


b(b^2-a^2
b(b-a)




=b-a

 

[Denis]

Close, but not quite; remember that b^2 - a^2 = (b + a)(b - a)

 

[chesterdg123]

Okay, then:

b(b^2-a^2)
--------------
b(b-a)

= b(b+a)(b-a)
b)(b-a)

=b+a

 

[Denis]

Yes.