Problems on Algebraic Fractions

Here we will learn how to simplify the problems on algebraic fractions to its lowest term.

1. Reduce the algebraic fractions to their lowest terms: $\frac{x^{2}  -  y^{2}}{x^{3}  -  x^{2}y}$

Solution:

$\frac{x^{2}  -  y^{2}}{x^{3}  -  x^{2}y}$

Factorizing the numerator and denominator separately and cancelling the common factors we get,

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

= $\frac{x  +  y}{x^{2}}$

2. Reduce to lowest terms $\frac{x^{2}  +  x  -  6}{x^{2}  -  4}$

Solution:

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

Step 1: Factorize the numerator x$^{2}$ + x – 6

                                         = x$^{2}$ + 3x – 2x – 6

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

                                         = (x + 3) (x – 2)

Step 2: Factorize the denominator: x$^{2}$ – 4

                                             = x$^{2}$ – 2$^{2}$

                                             = (x + 2) (x – 2)

Step 3: From steps 1 and 2: $\frac{x^{2}  +  x  -  6}{x^{2}  -  4}$

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

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

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

3. Simplify the algebraic fractions $\frac{36x^{2}  -  4}{9x^{2}  +  6x  +  1}$

Solution:

$\frac{36x^{2}  -  4}{9x^{2}  +  6x  +  1}$

Step 1: Factorize the numerator: 36x$^{2}$ – 4

                                           = 4(9x$^{2}$ – 1)

                                           = 4[(3x)$^{2}$ – (1)$^{2}$]

                                           = 4(3x + 1) (3x – 1)

Step 2: Factorize the denominator: 9x$^{2}$ + 6x + 1

                                             = 9x$^{2}$ + 3x + 3x + 1

                                             = 3x(3x + 1) + 1(3x + 1)

                                             = (3x + 1) (3x + 1)

Step 3: Simplification of the given expression after factorizing the numerator and the denominator:

$\frac{36x^{2}  -  4}{9x^{2}  +  6x  +  1}$

= $\frac{4(3x  +  1)(3x  -  1)}{(3x  +  1)(3x  +  1)}$

= $\frac{4(3x  -  1)}{(3x  +  1)}$

4. Reduce and simplify: $\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )$

Solution:

$\frac{8x^{3}y^{2}z}{2xy^{3}} of \left ( \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \div \frac{7xy^{2}}{35x^{2}yz^{3}}\right )$

= $\frac{8x^{3}y^{2}z}{2xy^{3}} of \frac{5x^{5}y^{2}z^{2}}{25xy^{3}z} \times \frac{35x^{2}yz^{3}}{7xy^{2}}$

= $\frac{4x^{3}y^{2}z}{xy^{3}} \left ( \frac{x^{5}y^{2}z^{2}}{xy^{3}z} \times \frac{x^{2}yz^{3}}{xy^{2}} \right )$

= 4x$^{10 - 3}$ ∙ y$^{-3}$ ∙ z$^{5}$

= $\frac{4x^{7}\cdot z^{5}}{y^{3}}$

5. Simplify: $\frac{2x^{2}  -  3x  -  2}{x^{2}  +  x  -  2} \div \frac{2x^{2}  +  3x  +  1}{3x^{2}  +  3x  -  6}$

Solution:

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

Step 1: First factorize each of the polynomials separately:

2x$^{2}$ – 3x – 2 = 2x$^{2}$ – 4x + x – 2

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

                 = (x – 2) (2x + 1)

x$^{2}$ + x – 2 = x$^{2}$ + 2x - x – 2

              = x(x + 2) - 1 (x + 2)

              = (x + 2) (x - 1)

2x$^{2}$ + 3x + 1 = 2x$^{2}$ + 2x + x + 1

                 = 2x(x + 1) + 1 (x + 1)

                 = (x + 1) (2x + 1)

3x$^{2}$ + 3x – 6 = 3[x$^{2}$ + x – 2]

                 = 3[x$^{2}$ + 2x - x – 2]

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

                 = 3[(x + 2) (x - 1)]

                 = 3[(x + 2) (x - 1)]

                 = 3(x + 2) (x - 1)

Step 2: Simplify the given expressions by substituting with their factors

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

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

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

= $\frac{3(x  -  2)}{(x  +  1)}$

8th Grade Math Practice

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