Question

Find the value of $x$ in the following equation using theorems on equal ratio.

$\frac{14x^2-6x+8}{10x^2+4x+7}=\frac{7x-3}{5x+2}$

• A. $\frac{37}{3}$
• B. $\frac{37}{9}$
• C. $\frac{37}{4}$
• D. $\frac{37}{5}$

Answer 1

The correct option is B $\frac{37}{9}$

Given:

$\frac{14x^2-6x+8}{10x^2+4x+7}=\frac{7x-3}{5x+2}$

Let, $\frac{14x^2-6x+8}{10x^2+4x+7}=\frac{7x-3}{5x+2}=k$
multiplying predecessor and successor of RHS by $-2x$ and LHS by 1. We get:

$\frac{14x^2-6x+8}{10x^2+4x+7}=\frac{-2x(7x-3)}{-2x(5x+2)}=k$
Using theorems of equal ratios:
$\frac{a}{b}=\frac{c}{d}=\frac{al+cm}{bl+dm}$

$\Rightarrow \frac{(14x^2-6x+8)+[-2x(7x-3\left)\right]}{(10x^2+4x+7)+[-2x(5x+2\left)\right]}=k$

$\Rightarrow \frac{(14x^2-6x+8)+(-14x^2+6x)]}{(10x^2+4x+7)+(-10x^2-4x)}=k$

$\Rightarrow k=\frac{8}{7}$

Now, using the LHS of the equation:

$\frac{7x-3}{5x+2}=k$

$\Rightarrow \frac{7x-3}{5x+2}=\frac{8}{7}$

$\Rightarrow 49x-21=40x+16$

$\Rightarrow 49x-40x=16+21$

$\Rightarrow 9x=37$

$\Rightarrow x=\frac{37}{9}$