Question

Find the value of $x$ is $\frac{1-x}{6}+\frac{2x}{3}-\frac{1-\frac{3}{7x}}{4}=2\frac{1}{6}$

Answer 1

$\frac{1-x}{6}+\frac{2x}{3}-\frac{1-3/7x}{4}=2\frac{1}{6}$

$\Rightarrow \frac{1-x}{6}+\frac{2x}{3}-\frac{7x-3}{28x}=\frac{13}{6}$

$\Rightarrow \frac{1-x+4x}{6}-\frac{7x-3}{28x}=\frac{13}{6}$

$\Rightarrow \frac{3x+1}{3}-\frac{7x-3}{14x}=\frac{13}{3}$

$\Rightarrow \frac{3x+1-13}{3}=\frac{7x-3}{14x}$

$\Rightarrow \frac{3x-12}{3}=\frac{7x-3}{14x}$

$\Rightarrow (x-4)\left(14x\right)=(7x-3)$

$\Rightarrow 14x^2-56x-7x+3=0$

$\Rightarrow 14x^2-63x+3=0$

$x=\frac{63\pm \sqrt{{63}^2-4\times 14\times 3}}{2\times 14}$

$=\frac{63\pm \sqrt{3969-168}}{28}$

$=\frac{63\pm \sqrt{3801}}{28}$

$=\frac{63\pm 61.65}{28}$

$=4.45$ or $0.048$.