Substitution in Fractions

Topic Content:

Example 6.4.1:

a. If m : n = 2 : 1, evaluate \( \frac{3m^2 \: -\: 2n^2}{m^2 \: + \: mn} \)

b. If \(\frac{x}{y} = \frac{2}{3}\scriptsize,\)evaluate \( \frac{3x \: + \: 2y}{x \: -\: \normalsize \frac{1}{3}y} \)

c. If x = \( \frac{3a \: – \: 2}{2a \: + \: 3}\scriptsize,\)express \( \frac{2x\:-\:1}{x\:+\:1}\)in terms of a

Solution

a. If m : n = 2 : 1, evaluate \( \frac{3m^2 \: -\: 2n^2}{m^2 \: + \: mn} \)

Step 1: m : n =2 : 1 (given)

⇒ \( \frac{m}{n} = \frac{2}{1} \)

m = 2n……….(1)

Step 2: Substituting equation 1 into the fraction

⇒ \( \frac{3m^2 \: -\: 2n^2}{m^2 \: + \: mn} \)

⇒ \( \frac{3(2n)^2 \: -\: 2n^2}{(2n)^2 \: + \: (2n)n} \)

⇒ \( \frac{12n^2 \: -\: 2n^2}{4n^2 \: + \: 2n^2} \)

= \( \frac{10n^2}{6n^2} \)

= \( \frac{5}{3} \)

b. If \(\frac{x}{y} = \frac{2}{3}\scriptsize,\)evaluate \( \frac{3x \: + \: 2y}{x \: -\: \normalsize \frac{1}{3}y} \)

Step 1: Divide the numerator and denominator by y

⇒ \(\large \frac{\frac{3x}{y} \normalsize \: + \: 2}{\frac{x}{y} \: – \: \frac{1}{3}} \)

Step 2: Substitute \( \frac{2}{3}\)for \( \frac{x}{y}\)

⇒ \(\frac{\normalsize 3\large \frac{x}{y}\normalsize \: + \: 2}{ \large \frac{x}{y} \: – \: \frac{1}{3}} \)

⇒ \(\frac{\normalsize 3 \: \times \: \large \frac{2}{3} \normalsize \: + \: 2}{\large \frac{2}{3} \: – \: \frac{1}{3}} \)

⇒ …