Multiplication of Algebraic Fractions

To solve the problems on multiplication of algebraic fractions we will follow the same rules that we already learnt in multiplication of fractions in arithmetic.

From multiplication of fractions we know,

Product of two or more fractions = $$\frac{Product of numerators}{Product of denominators}$$

In algebraic fractions, the product of two or more fractions can be determined in the same way i.e.

Product of two or more fractions = $$\frac{Product of numerators}{Product of denominators}$$.

1. Determine the product of the following algebraic fractions:

(i) $$\frac{m}{n} \times \frac{a}{b}$$

Solution:

$$\frac{m}{n} \times \frac{a}{b}$$

=  $$\frac{m \cdot a}{n \cdot b}$$

= $$\frac{am}{bn}$$

(ii) $$\frac{x}{x + y} \times \frac{y}{x - y}$$

Solution:

$$\frac{x}{x + y} \times \frac{y}{x - y}$$

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

= $$\frac{xy}{x^{2} - y^{2}}$$

2. Find the product of the algebraic fractions in the lowest form: $$\frac{m}{p + q} \times \frac{m}{n} \times \frac{n(p - q)}{m(p + q)}$$

Solution:

$$\frac{m}{p + q} \times \frac{m}{n} \times \frac{n(p - q)}{m(p + q)}$$

= $$\frac{m \cdot m \cdot n(p - q)}{(p + q) \cdot n \cdot m(p + q)}$$

= $$\frac{m^{2}n(p - q)}{mn(p + q)^{2}}$$

Here the numerator and denominator have a common factor mn, so by dividing the numerator and denominator of the product by mn, the product in the lowest form will be $$\frac{m (p - q)}{(p + q)^{2}}$$.

3. Find the product and express in the lowest form: $$\frac{x(x + y)}{x - y} \times \frac{x - y}{y(x + y)} \times \frac{x}{y}$$

Solution:

$$\frac{x(x + y)}{x - y} \times \frac{x - y}{y(x + y)} \times \frac{x}{y}$$

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

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

Here, the common factor in the numerator and denominator is (x + y) (x – y). If the numerator and denominator are divided by this common factor, the product in the lowest form will be $$\frac{x^{2}}{y^{2}}$$.

4. Find the product of the algebraic fraction: $$\left ( \frac{5a}{2a - 1} - \frac{a - 2}{a} \right ) \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$

Solution:

$$\left ( \frac{5a}{2a - 1} - \frac{a - 2}{a} \right ) \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$

Here, the L.C.M. of the denominators of the first part is a(2a – 1) and the L.C.M. of the denominators of the second part is (a + 2)

Therefore,  $$\left \{\frac{5a \cdot a}{(2a - 1) \cdot a} - \frac{(a - 2) \cdot (2a - 1)}{a \cdot (2a - 1)} \right \} \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$

= $$\{ \frac{5a^{2}}{a(2a - 1)} - \frac{(a - 2)(2a - 1)}{a(2a - 1)} \} \times \left ( \frac{2a}{a + 2} - \frac{1}{a + 2}\right )$$

= $$\frac{5a^{2} - (a - 2)(2a - 1)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

= $$\frac{5a^{2} - (2a^{2} - 5a + 2)}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

= $$\frac{5a^{2} - 2a^{2} + 5a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

= $$\frac{3a^{2} + 5a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

= $$\frac{3a^{2} + 6a - a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

= $$\frac{3a^{2} + 6a - a - 2}{a(2a - 1)} \times \frac{2a - 1}{a + 2}$$

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

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

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

Here, the common factor in the numerator and denominator is (x + 2) (2x - 1). If the numerator and denominator are divided by this common factor, the product in the lowest form will be

= $$\frac{(3a - 1)}{a}$$

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