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Value of an Expression

Solve 12 × 2+...

Question

Solve $\frac{12 x^{2} + 18 x + 12}{18 x^{2} + 12 x + 58} = \frac{2 x + 3}{3 x + 2}$

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Solution

$Let : \frac{12 x^{2} + 18 x + 12}{18 x^{2} + 12 x + 58} = \frac{2 x + 3}{3 x + 2} = k (x \neq 0) If we multiply both the terms of the second ratio by 6 x, we would obtain the first two terms in each of the antecedents and consequents of the first ratio .$

$On multiplying both the terms of the second ratio by 6 x, we get : \frac{12 x^{2} + 18 x + 12}{18 x^{2} + 12 x + 58} = \frac{6 x (2 x + 3)}{6 x (3 x + 2)} = k By the theorem on equal ratios, we get : \frac{12 x^{2} + 18 x + 12 - 6 x (2 x + 3)}{18 x^{2} + 12 x + 58 - 6 x (3 x + 2)} = k \Rightarrow \frac{12 x^{2} + 18 x + 12 - 12 x^{2} - 18 x}{18 x^{2} + 12 x + 58 - 18 x^{2} - 12 x} = k \Rightarrow \frac{12}{58} = k Now, \frac{2 x + 3}{3 x + 2} = \frac{12}{58} \Rightarrow 116 x + 174 = 36 x + 24 \Rightarrow 80 x = - 150 \Rightarrow x = \frac{- 15}{8}$

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\frac{12 x^{2} + 18 x + 42}{18 x^{2} + 12 x + 58} = \frac{2 x + 3}{3 x + 2}

\frac{12 x^{2} + 18 x + 12}{18 x^{2} + 12 x + 58} = \frac{2 x + 3}{3 x + 2}

- \frac{15}{8}

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