# If 1 is added to both the numerator and denominator of a fraction its value becomes 1/2 and if 1 subtracted from both the numerator and denominator of a fraction it becomes 1/3 then What is the original fraction?

Jan 10, 2016

3/7

#### Explanation: Feb 11, 2018

The original fraction is $\frac{3}{7}$

#### Explanation:

Let $\frac{x}{y}$ represent the fraction

The problem specifies

$\left(a\right)$ Adding $1$ to the numerator and denominator makes the fraction equal to $\frac{1}{2}$
$\left(b\right)$ Subtracting $1$ from the top and bottom makes the fraction equal to $\frac{1}{3}$

(a)   (x + 1)/(y+1) = (1)/(2)

(b)   (x - 1)/(y-1) = (1)/(3)

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots}$―――――――

Solve for $x$, already defined as "the numerator of the original equation"

These equations can be solved as Ratio&Proportion problems.

1) Cross multiply both equations
$\left(a\right)$   $2 \left(x + 1\right) = 1 \left(y + 1\right)$
$\left(b\right)$   3(x−1)=1(y−1)

2) Clear the parentheses
$\left(a\right)$   $2 x + 2 = y + 1$
$\left(b\right)$   $3 x - 3 = y - 1$

3) Write the equations with the variables on the left and the numbers on the right
$\left(a\right)$   $2 x - y =$ - $1$
$\left(b\right)$   3x - y =     2

4) Start with Equation $\left(b\right)$ and subtract Equation $\left(a\right)$
(Do it in this order to avoid negative numbers.)
$\left(b\right)$      3x - y =   2
$\left(a\right)$ - ($2 x - y =$ - $1$)

5) Clear the parentheses and combine to eliminate the $y$ term
$\left(b\right)$     $3 x - y = 2$
$\left(a\right)$ $- 2 x + y = 1$
―――――――――
$\textcolor{w h i t e}{\ldots \ldots \ldots .} x$ $\textcolor{w h i t e}{\ldots m} = 3$ $\leftarrow$ already defined as "the numerator of the original fraction"

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots}$―――――――

Solve for $y$, already defined as "the denominator of the original fraction"

1) Using one of the original equations, sub in $3$ in the place of $x$
$\frac{\textcolor{b l u e}{x} - 1}{y - 1} = \frac{1}{3}$

$\frac{\textcolor{b l u e}{3} - 1}{y - 1} = \frac{1}{3}$

2) Cross multiply and solve for $y$, already defined as "the denominator of the original fraction"
$3 \left(3 - 1\right) = 1 \left(y - 1\right)$

3) Solve inside the parentheses
$3 \left(2\right) = 1 \left(y - 1\right)$

4) Clear the parentheses by distributing the $3$ and the $1$
$6 = y - 1$

5) Add $1$ to both sides to isolate $y$

6) $7 = y$ $\leftarrow$ answer for $y$, defined as "the denominator of the original fraction"

$\text{Answer}$
The original fraction is $\frac{3}{7}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots}$―――――――

Check
Add $1$ to the top and bottom of $\frac{3}{7}$ to see if it becomes $\frac{1}{2}$

$\frac{3 + 1}{7 + 1} =$ $\frac{4}{8} = \frac{1}{2}$  ✓

Subtract $1$ from the top and bottom to see if it becomes $\frac{1}{3}$

$\frac{3 - 1}{7 - 1} =$ $\frac{2}{6} = \frac{1}{3}$  ✓

$C h e c k$