Algebra II Review: Solutions To Algebraic Fraction Problems

 Simplify each of the following:

1) x²  - y²/ y²  - x²

Solution:

Factor numerator and denominator:

(x + y) (x - y)/ (y + x) (y - x)

Change sign of (y - x) in denominator to - (x - y):

Þ  

(x + y) (x - y)/ - (x - y) (y + x)

We can then cancel the (x - y)  factors in numerator and denominator:

=  -  (x + y) / (x + y)  =   -1

2)  27 a³ x²y ⁵  / 36 a ² x⁴y⁴

^Solution:  

^{Divide, first reducing 27/36 to 3/4 and paying attention to properties of exponents (the realm of Algebra I)}

^{3 a y  / 4 x2}

3) 2a ^1/2 (x + y)/  5 x a ^3/2

Solution:  

Apply quotient rule for exponents (from Algebra I):

a ^m /a ⁿ     =  a ^{m - n}

In this case, m = 1/2 and n = 3/2 so m - n =  1/2 - 3/2 =  -1

So:  a^{- 1}  =  1/ a

Then the fraction simplifies to: 

2 (x + y)/ 5 a x

4) (x -1) (2 - x) (3 - x)/ (1 - x) (x - 2) (x - 4)

Solution:  

Notice that the numerator contains (x - 1) and (2 - x), while the denominator contains (1 - x) and (x - 2). This indicates application of sign rule.  So, following the process in initial post, we rewrite the factors in the numerator to match those in denominator by factoring out (-1) in each, viz.

(x -1)  =  (-1) (1 - x)

(2 - x) =   (-1) (x - 2)

Substituting back into the numerator we obtain:

(1 - x) (x - 2) (3 - x)   (Rem: (-1) (-1) = +1)

So the algebraic fraction becomes:

(1 - x) (x - 2) (3 - x) /  (1 - x) (x - 2) (x - 4)

After canceling common factors:

(3 - x) / (x - 4)