Question

Form the pair of linear equations for the following problems and find their solution by substitution method:

A fraction becomes $\frac{9}{11}$, if $2$ is added to both the numerator and the denominator. If, $3$ is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.

Answer 1

Step 1: Assume variables and form one equation

Let, the numerator $=x$

And, the denominator $=y$

So, the fraction $=\frac{x}{y}$

Now, according to the question,

$\frac{\mathrm{numerator}+2}{\mathrm{denominator}+2}=\frac{9}{11}$

$\Rightarrow$ $\frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow$ $11\left(x+2\right)=9\left(y+2\right)$ [cross multiplication]

$\Rightarrow$ $11x+22=9y+18$

$\Rightarrow$ $11x=9y+18-22$

$\Rightarrow$ $11x=9y-4$

$\Rightarrow$ $x=\frac{9y-4}{11}$ …(i)

Step 2: Form the second equation

Again, according to the question,

$\frac{\mathrm{numerator}+3}{\mathrm{denominator}+3}=\frac{5}{6}$

$\Rightarrow$ $\frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow$ $6\left(x+3\right)=5\left(y+3\right)$ [cross multiplication]

$\Rightarrow$ $6x+18=5y+15$

$\Rightarrow$ $6x=5y+15-18$

$\Rightarrow$ $6x=5y-3$ …(ii)

Step 3: Calculate $y$ by substitution

Substituting the value of $x$ in equation (ii) from equation (i),

$6\left(\frac{9y-4}{11}\right)=5y-3$

$\Rightarrow$ $\frac{6\left(9y-4\right)}{11}=5y-3$

$\Rightarrow$ $\frac{54y-24}{11}=5y-3$

$\Rightarrow$ $54y-24=11\left(5y-3\right)$

$\Rightarrow$ $54y-24=55y-33$

$\Rightarrow$ $33-24=55y-54y$

$\Rightarrow$ $9=y$

Or, $y=9$

Step 4: Calculate $x$ by substitution

Substituting the value of $y$ in equation (i),

$\because$ $x=\frac{9y-4}{11}$

$\Rightarrow$ $x=\frac{9\times 9-4}{11}$

$\Rightarrow$ $x=\frac{81-4}{11}$

$\Rightarrow$ $x=\frac{77}{11}$

$\Rightarrow$ $x=7$

So, the fraction $=\frac{x}{y}=\frac{7}{9}$

Hence, the required fraction is $\frac{7}{9}$.