Question
Solve for x: $\frac{3 x^2+5 x+18}{5 x^2+6 x+12}=\frac{3 x+5}{5 x-6}$.
✓
Answer
Multiplying the Numerator and Denominator of R.H.S. by -x
$\frac{3 x^2+5 x+18}{5 x^2+6 x+12}=\frac{-3 x^2+5}{-5 x^2-6}$
Since, each ratio $=\frac{\text { Sum of antecedents }}{\text { Sum of consequents }}$
So $\frac{3 x^2+5 x+18-3 x^2-5 x}{5 x^2+6 x+12-5 x^2-6 x}$
$=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \frac{18}{12}=\frac{-3 x^2-5 x}{-5 x^2-6 x}$
$ \Rightarrow \frac{3}{2}=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \Rightarrow \frac{3}{2}=\frac{3 x+5}{5 x+6} $
$\Rightarrow 15 x+18=+x+10$
$ \Rightarrow 9 x=-8$
$ \Rightarrow x=\frac{-8}{9} .$
$\frac{3 x^2+5 x+18}{5 x^2+6 x+12}=\frac{-3 x^2+5}{-5 x^2-6}$
Since, each ratio $=\frac{\text { Sum of antecedents }}{\text { Sum of consequents }}$
So $\frac{3 x^2+5 x+18-3 x^2-5 x}{5 x^2+6 x+12-5 x^2-6 x}$
$=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \frac{18}{12}=\frac{-3 x^2-5 x}{-5 x^2-6 x}$
$ \Rightarrow \frac{3}{2}=\frac{-3 x^2-5 x}{-5 x^2-6 x} $
$ \Rightarrow \frac{3}{2}=\frac{3 x+5}{5 x+6} $
$\Rightarrow 15 x+18=+x+10$
$ \Rightarrow 9 x=-8$
$ \Rightarrow x=\frac{-8}{9} .$
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