Question

The numerator of a fraction is $3$ less than its denominator. If $2$ is added to both the numerator and denominator, then the sum of the new fraction and the original fraction is $\dfrac{29}{20}$. Find the original fraction.

Answer 1

Hint:Assume either the numerator or denominator to be an unknown quantity. Then, use the condition given to solve for the unknown quantity, and thus form the original fraction.

Complete step-by-step answer:
Let us assume the denominator of the original fraction to be x. Now, the numerator is said to be 3 less than the denominator. Thus, the numerator is (x – 3).
Then, the original fraction is of the expression $\dfrac{\text{x - 3}}{\text{x}}$.
Now, by adding 2 to each of the numerator and denominator of the original fraction, we get,
The new fraction $=\text{ }\dfrac{\text{x - 1}}{\text{x + 2}}$
According to the condition i.e the sum of the new fraction and the original fraction is $\dfrac{29}{20}$ given in the question, the equation can be expressed as follows:
\[\begin{align}
  & \dfrac{\text{x }-\text{ 3}}{\text{x}}\text{ + }\dfrac{\text{x }-\text{ 1}}{\text{x }+\text{ 2}}\text{ = }\dfrac{29}{20} \\
 & \Rightarrow \text{ 1 }-\text{ }\dfrac{3}{\text{x}}\text{ }+\text{ }\dfrac{\text{x }-\text{ 1}}{\text{x + 2}}\text{ = }\dfrac{29}{20} \\
 & \Rightarrow \text{ }\dfrac{\text{x }-\text{ 1}}{\text{x + 2}}\text{ }-\text{ }\dfrac{3}{\text{x}}\text{ = }\dfrac{9}{20} \\
 & \Rightarrow \text{ }\dfrac{\text{x(x }-\text{ 1) }-\text{ 3(x + 2)}}{\text{x(x + 2)}}\text{ = }\dfrac{9}{20} \\
 & \Rightarrow \text{ }\dfrac{{{\text{x}}^{2}}\text{ }-\text{ x }-\text{ 3x }-\text{ 6}}{{{\text{x}}^{2}}\text{ + 2x}}\text{ = }\dfrac{9}{20} \\
 & \Rightarrow \text{ 20(}{{\text{x}}^{\text{2}}}\text{ - 4x - 6) = 9(}{{\text{x}}^{\text{2}}}\text{ + 2x)} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \text{ 20}{{\text{x}}^{2}}\text{ - 80x - 120 = 9}{{\text{x}}^{2}}\text{ + 18x} \\
 & \Rightarrow \text{ 11}{{\text{x}}^{2}}\text{ - 98x - 120 = 0} \\
 & \Rightarrow \text{ 11}{{\text{x}}^{2}}\text{ - 110x + 12x - 120 = 0} \\
 & \Rightarrow \text{ 11x(x - 10) + 12(x - 10) = 0} \\
 & \Rightarrow \text{ (x - 10)(11x + 12) = 0} \\
\end{align}\]
Either $\text{x = 10}$ or $\text{x }=\text{ }-\frac{12}{11}$
Thus, the solutions of the original fraction are given as:
i) $\dfrac{10\text{ - 3}}{10}\text{ = }\dfrac{7}{10}$
 ii) $\dfrac{-\dfrac{12}{11}\text{ }-\text{ 3}}{-\dfrac{12}{11}}\text{ = }\dfrac{\dfrac{-12\text{ }-\text{ 33}}{11}}{-\dfrac{12}{11}}$
$\begin{align}
  & \text{ }=\text{ }\dfrac{\text{ }-\dfrac{45}{11}}{-\dfrac{12}{11}} \\
 & \text{ }=\text{ }\dfrac{15}{4} \\
\end{align}$
Since, in the solution $\dfrac{15}{4}$, the condition given in the question that the numerator is 3 less than the denominator, is not satisfied, this solution is discarded. So, only the solution $\dfrac{7}{10}$ is accepted.
Thus, the answer is$\dfrac{7}{10}$.

Note: Of the two solutions for the original fraction, the solution$\dfrac{7}{10}$, both the conditions given in the question are satisfied. But, the solution $\dfrac{15}{4}$ satisfies neither of the given conditions. It is obtained by reducing the fraction $\dfrac{-\dfrac{45}{11}}{-\dfrac{12}{11}}$ into its lowest terms, the original solution $\dfrac{-\dfrac{45}{11}}{-\dfrac{12}{11}}$ satisfies both the conditions.