How to Simplify Rational Expressions with Variables

[captainnumber36]

Problem 1:

1 / (2x-y) - 2 / (x+2y) = ?

The answer is:

(4y-3x) / (2x^2+3xy-2y^2)

Please explain.Problem 2:

f(x) = 4x/(1-x) and g(x) - 2/x, then f(g(x)) = ?

The answer is 8/(x-2)

Please explain.Problem 3:

Square Root of a / (1+ Square Root of a) = ?

The answer is (Square Root of a - a) / (1-a)

Please explain.Thank you very much in advance.

 

[captainnumber36]

Here is a photo of the problems if it helps: 18-20.

https://imgur.com/a/Jeis8T4

 

[MountEvariste]

Can I ask you what $\displaystyle \frac{4}{8-4}$ is? By your reasoning for the answer to these equations, it's:

$\displaystyle \frac{4}{8-4}= \frac{\cancel{4}}{8-\cancel{4}} = \frac{1}{8}.$ But you know that $\displaystyle \frac{4}{8-4} = \frac{4}{4} = 1.$

When doing algebraic fractions, don't do anything you wouldn't do for numerical fractions.

 

[HOI]

The problem is to find [math]\frac{1}{2x- y}- \frac{2}{x+ 2y}[/math].

To add or subtract fractions, we need to get a "common denominator". Here the common denominator is (2x- y)(x+ 2y).

Multiply numerator and denominator of the first fraction by x+ 2y:
[math]\frac{1}{2x- y}\frac{x+ 2y}{x+ 2y}= \frac{x+ 2y}{(2x- y)(x+ 2y)}[/math].

Multiply numerator and denominator of the second fraction by 2x- y:
[math]\frac{2}{x+ 2y}\frac{2x- y}{2x- y}= \frac{4x- 2y}{(2x- y)(x+ 2y)}[/math].

Now we are ready to subtract the fractions:
[math]\frac{x+ 2y}{(2x- y)(x+ 2y)}- \frac{4x- 2y}{(2x- y)(x+ 2y)}= \frac{-3x+ 4y}{(2x- y)(x+ 2y)}[/math].

The given possible answers do not have the denominator factored so calculate [math](2x- y)(x+ 2y)= 2x(x+ 2y)- y(x+ 2y)= 2x^2+ 4xy- xy- 2y^2= 2x^2+ 3xy- 2y^2[/math].

The fraction is [math]\frac{4y- 3x}{2x^2+ 3xy- 2y^2}[/math]. That is answer "E".

To simplify [math]\frac{\sqrt{a}}{1+ \sqrt{a}}[/math], "rationalize the denominator". You should have learned earlier that [math](a+ b)(a- b)= a(a- b)+ b(a- b)= a^2- ab+ ab- b^2= a^2- b^2[/math] since the "ab" terms cancel. The "[math]a^2[/math]" and "[math]b^2[/math]" terms get rid of the square roots.

Here the "a+ b" is [math]1+ \sqrt{a}[/math]. Multiply both numerator and denominator of [math]\frac{\sqrt{a}}{1+ \sqrt{a}}[/math] by [math]1- \sqrt{a}[/math]: [math]\frac{\sqrt{a}}{1+ \sqrt{a}}\frac{1- \sqrt{a}}{1- \sqrt{a}}[/math][math]= \frac{\sqrt{a}- \sqrt{a}^2}{1^2- \sqrt{a}^2}=[/math][math] \frac{\sqrt{a}- a}{1- a}[/math].

That is answer "B".

 

[captainnumber36]

Thanks to both of you! :)