Simplification of Algebraic Fractions

Here we will learn simplification of algebraic fractions to its lowest term.

1. Simplify the algebraic fraction:

$\frac{8a^{2}b}{4a^{2}  +  6ab}$

Solution:

$\frac{8a^{2}b}{4a^{2}  +  6ab}$

We see in the given fraction the numerator is monomial and the denominator is binomial, which can be factorized.

= $\frac{\not{2}\times 2\times 2\times \not{a}\times a\times b}{\not{2}\not{a}(2a  +  3b)}$

We can see that ‘2’ and ‘a’ are the common factors in the numerator and denominator so, we cancel the common factor ‘2’ and ‘a' from the numerator and denominator.

= $\frac{4ab}{(2a  +  3b)}$

2. Reduce the algebraic fraction to its lowest term:

$\frac{x^{2}  +  8x  +  12}{x^{2}  -  4}$

Solution:

$\frac{x^{2}  +  8x  +  12}{x^{2}  -  4}$

Each of the numerator and denominator is polynomial, which can be factorized.

= $\frac{x^{2}  +  6x  +  2x  +  12}{(x)^{2}  -  (2)^{2}}$

 = $\frac{x(x  +  6 )  +  2(x  +  6)}{(x  +  2)(x  -  2)}$

= $\frac{(x  +  2)(x  +  6)}{(x  +  2)(x  -  2)}$

We observed that in the numerator and denominator (x + 2) is the common factor and there is no other common factor. Now, we cancel the common factor from the numerator and denominator.

= $\frac{(x  +  6)}{(x  -  2)}$

3. Reduce the algebraic fraction to its lowest form:

$\frac{5x^{2}  -  45}{x^{2}  -  x  -  12}$

Solution:

$\frac{5x^{2}  -  45}{x^{2}  -  x  -  12}$

Each of the numerator and denominator is polynomial, which can be factorized.

= $\frac{5(x^{2}  -  9)}{x^{2}  -  4x  +  3x  -  12}$

= $\frac{5[(x)^{2}  -  (3)^{2}]}{x(x  -  4)  +  3(x  -  4)}$

= $\frac{5(x  +  3)(x  -  3)}{(x  +  3)(x  -  4)}$

Here, in the numerator and denominator (x + 3) is the common factor and there is no other common factor. Now, we cancel the common factor from the numerator and denominator.

= $\frac{5(x  -  3)}{(x  -  4)}$

4. Simplify the algebraic fraction:

$\frac{x^{4}  -  13x^{2}  +  36}{2x^{2}  +  10x  +  12}$

Solution:

$\frac{5x^{2}  -  45}{x^{2}  -  x  -  12}$

Each of the numerator and denominator is polynomial, which can be factorized.

= $\frac{x^{4}  -  9x^{2}  -  4x^{2}  +  36}{2(x^{2}  +  5x  +  6)}$

= $\frac{x^{2}(x^{2}  -  9)  -  4(x^{2}  -  9)}{2(x^{2}  +  2x  +  3x  +  6)}$

= $\frac{(x^{2}  -  4)(x^{2}  -  9)}{2[x(x  +  2)  +  3(x  +  2)]}$

= $\frac{(x^{2}  -  4)(x^{2}  -  9)}{2(x  +  2)(x  +  3)} [Since, a^{2}  -  b^{2} = (a  +  b)(a  -  b)]$

= $\frac{(x  +  2)(x  -  2)(x  +  3)(x  -  3)}{2(x  +  2)(x  +  3)}$

Here, in the numerator and denominator (x + 2) and (x + 3) are the common factors and there is no other common factor. Now, we cancel the common factors from the numerator and denominator.

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

5. Reduce the algebraic fraction to its lowest term:

$\frac{x^{2}  +  5x  -  2}{2x^{2}  +  x  -  6} \div \frac{4x^{2}  -  9}{6x^{2}  +  7x  -  3}$

Solution:

$\frac{x^{2}  +  5x  -  2}{2x^{2}  +  x  -  6} \div \frac{4x^{2}  -  9}{6x^{2}  +  7x  -  3}$

Each of the numerator and denominator of each fraction are polynomial, which can be factorized.

Now by factorizing each polynomial we get;

3x² + 5x – 2 = 3x² –x + 6x – 2

                 = 3(3x – 1) + 2(3x – 1)

                 = (x + 2)(3x – 1)

2x² + x – 6 = 2x² - 3x - 4x - 6

                = x(2x – 3) + 2(2x – 3)

                = (x + 2)(2x - 3)

4x² – 9 = (2x)² - (3)²

           = (2x + 3)(2x – 3)

6x² + 7x – 3 = 6x² – 2x + 9x – 3

                  = 2x(3x – 1) + 3(3x – 1)

                  = (2x + 3)(3x – 1)

Therefore, we have

$\frac{(x  +  2)(3x  -  1)}{(x  +  2)(2x  -  3)} \div \frac{(2x  +  3)(2x  -  3)}{(2x  +  3)(3x  -  1)}$

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

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

= $\frac{9x^{2}  -  6x  +  1}{4x^{2}  -  12x  +  9}$

 

6. Reduce the algebraic fraction to its lowest form:

 $\frac{1}{x^{2}  -  3x  +  2}  +  \frac{1}{x^{2}  -  5x  +  6}  +  \frac{1}{x^{2}  -  4x  +  3}$

Solution:

$\frac{1}{x^{2}  -  3x  +  2}  +  \frac{1}{x^{2}  -  5x  +  6}  +  \frac{1}{x^{2}  -  4x  +  3}$

= $\frac{1}{x^{2}  -  2x  -  x  +  2}  +  \frac{1}{x^{2}  -  3x  -  2x  +  6}  +  \frac{1}{x^{2}  -  x  -  3x  +  3}$

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

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

= $\frac{1 \times (x  -  3)}{(x  -  2)(x  -  1)(x  -  3)}  +  \frac{1\times (x  -  1)}{(x  -  3)(x  -  2)(x  -  1)}  +  \frac{1\times (x  -   2)}{(x  -  1)(x  -  3)(x  -  2)}$

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

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

= $\frac{(3x  -  6)}{(x  -  1)(x  -  2)(x  -  3)}$

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

= $\frac{3}{(x  -  1)(x  -  3)}$

 

7. Simplify the algebraic fraction:

$\frac{3x}{x  -  2}  +  \frac{5x}{x^{2}  -  4}$

Solution:

$\frac{3x}{x  -  2}  +  \frac{5x}{x^{2}  -  4}$

= $\frac{3x}{x  -  2}  +  \frac{5x}{x^{2}  -  (2)^{2}}$

= $\frac{3x}{x  -  2}  +  \frac{5x}{(x  +  2)(x  -  2)}$

= $\frac{3x \times (x  +  2)}{(x  -  2)(x  +  2)}  +  \frac{5x}{(x  +  2)(x  -  2)}$

= $\frac{3x(x  +  2)  -  5x}{(x  -  2)(x  +  2)}$

= $\frac{3x^{2}  +  6x  -  5x}{(x  -  2)(x  +  2)}$

= $\frac{3x^{2}  +  x}{(x  -  2)(x  +  2)}$

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

8th Grade Math Practice

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