Question

A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get $\frac{8}{11}$, but if the numerator is increased by 8 and the denominator is doubled, we get $\frac{2}{5}$. Find the fraction.

Answer 1

Let the fraction be $\frac{x}{y}$. According to the given conditions, we have
$\frac{3x}{y-3}=\frac{18}{11}$ and $\frac{x+8}{2y}=\frac{2}{3}$
$\to 11x=6y-18$ and $5x+40=4y$
So, we have
$11x-6y+18=0$ (1)
$5x-4y+40=0$ (2)
On comparing the coefficients of (1) and (2) with $a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$
we have $a_1=11,b_1=-6,c_1=18;a_2=5,b_2=-4,c_2=40$
Thus, $A_1{b}_2-a_2{b}_1=\left(11\right)(-4)-\left(5\right)(-6)=-14\ne 0$.
Hence, the system has a unique solution. Now, writing the coefficients for the cross multiplication, we have
$\to \frac{x}{-240+72}=\frac{y}{90-440}=\frac{1}{-44+30}$
$\to \frac{x}{-168}=\frac{y}{-350}=\frac{1}{-14}$
Thus,
$x=\frac{168}{14}=12;y=\frac{350}{14}=25$.
Hence, the fraction is $\frac{12}{25}$.