Question

solve it in detail
Find the value of x​

Answer 1

$\sf \Large Solution!!$

$\sf Let\:\dfrac{12x^{2}+18x+42}{18x^{2}+12x+58}=\dfrac{2x+3}{3x+2}=A\dots (i)$

Multiply the numerator and denominator of second ratio by 6x.

$\sf \to A=\dfrac{12x^{2}+18x+42}{18x^{2}+12x+58}=\dfrac{6x(2x+3)}{6x(3x+2)}$

$\sf \to A=\dfrac{12x^{2}+18x+42}{18x^{2}+12x+58}=\dfrac{12x^{2}+18x}{18x^{2}+12x}$

Theorem on equal ratios.

$\sf \to A=\dfrac{(12x^{2}+18x+42)-(12x^{2}+18x)}{(18x^{2}+12x+58)-(18x^{2}+12x)}$

$\sf \to A=\dfrac{\cancel{12x^{2}}\cancel{+18x}+42\cancel{-12x^{2}}\cancel{-18x}}{\cancel{18x^{2}}\cancel{+12x}+58\cancel{-18x^{2}}\cancel{-12x}}$

$\sf \to A=\dfrac{42}{58}=\dfrac{21}{29}$

From (i)

$\sf \to \dfrac{2x+3}{3x+2}=A$

$\sf \to \dfrac{2x+3}{3x+2}=\dfrac{21}{29}$

Cross-multiplication.

$\sf \to 29(2x+3)=21(3x+2)$

$\sf \to 58x+87=63x+42$

$\sf \to 58x-63x=42-87$

$\sf \to \cancel -5x=\cancel -45$

$\sf \to x=\dfrac{45}{5}$

$\sf \to x=9$