# 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

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

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

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

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