Question

Solve :

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

Answer 1

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

If x = 0, then $\frac{12\times 0+18\times 0+42}{18\times 0+12\times 0+58}=\frac{2\times 0+3}{3\times 0+2}\Rightarrow \frac{42}{58}=\frac{3}{2}$, which is not true.

So, x = 0 is not a solution of the given equation.

Now,

$$\begin{gathered} \frac{12x^2+18x+42}{18x^2+12x+58}=\frac{2x+3}{3x+2}=\frac{6x\left(2x+3\right)}{6x\left(3x+2\right)} \\ \Rightarrow \frac{12x^2+18x+42}{18x^2+12x+58}=\frac{2x+3}{3x+2}=\frac{\left(12x^2+18x+42\right)-6x\left(2x+3\right)}{\left(18x^2+12x+58\right)-6x\left(3x+2\right)}\left(\mathrm{Theorem}\mathrm{of}\mathrm{equal}\mathrm{ratios}\right) \\ \Rightarrow \frac{12x^2+18x+42}{18x^2+12x+58}=\frac{2x+3}{3x+2}=\frac{12x^2+18x+42-12x^2-18x}{18x^2+12x+58-18x^2-12x} \\ \Rightarrow \frac{12x^2+18x+42}{18x^2+12x+58}=\frac{2x+3}{3x+2}=\frac{42}{58} \end{gathered}$$
$$\begin{gathered} \therefore \frac{2x+3}{3x+2}=\frac{42}{58} \\ \Rightarrow \frac{2x+3}{3x+2}=\frac{21}{29} \\ \Rightarrow 58x+87=63x+42 \\ \Rightarrow 63x-58x=87-42 \\ \Rightarrow 5x=45 \\ \Rightarrow x=9 \end{gathered}$$
Thus, the solution of the given equation is x = 9.