Question

solve this problem . When
A fraction becomes 1/3 if 2 is added to both numerator and denominator. If 3 is added to both numerator and denominator it becomes 2/5. Assuming the original fraction to be x/y, form a pair of linear equations in two variables for the problem.​

Answer 1

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Answer 2

Answer

Let the numerator be x.

Let the denominator be y.

So, the fraction becomes

$\sf \dashrightarrow \dfrac{x}{y}$

According to the first statement,

The fraction becomes 1/3 if 2 is added to both numerator and denominator. So,

$\sf \dashrightarrow \dfrac{x + 2}{y + 2} = \dfrac{1}{3}$

$\sf \dashrightarrow 3(x + 2) = 1(y + 2)$

$\sf \dashrightarrow 3x + 6 = y + 2$

$\sf \dashrightarrow 3x - y = 2 - 6$

$\sf \dashrightarrow 3x - y = -4 \: \: --- (i)$

According to the second statement,

The fraction becomes 2/5 if 3 is added to both numerator and denominator. So,

$\sf \dashrightarrow \dfrac{x + 3}{y + 3} = \dfrac{2}{5}$

$\sf \dashrightarrow 5(x + 3) = 2(y + 3)$

$\sf \dashrightarrow 5x + 15 = 2y + 6$

$\sf \dashrightarrow 5x - 2y = 6 - 15$

$\sf \dashrightarrow 5x - 2y = -9 \: \: --- (ii)$

Now, by first equation,

$\sf \dashrightarrow 3x - y = -4$

$\sf \dashrightarrow 3x = -4 + y$

$\sf \dashrightarrow x = \dfrac{-4 + y}{3}$

Now, let's find the value of y by second equation.

$\sf \dashrightarrow 5x - 2y = -9$

$\sf \dashrightarrow 5 \bigg( \dfrac{-4 + y}{3} \bigg) - 2y = -9$

$\sf \dashrightarrow \dfrac{-20 + 5y}{3} - 2y = -9$

$\sf \dashrightarrow \dfrac{-20 + 5y - 6y}{3} = -9$

$\sf \dashrightarrow \dfrac{-20 - 1y}{3} = -9$

$\sf \dashrightarrow -20 - 1y = -9 \times 3$

$\sf \dashrightarrow -20 - 1y = -27$

$\sf \dashrightarrow -1y = -27 + 20$

$\sf \dashrightarrow -1y = -7$

$\sf \dashrightarrow y = 7$

Now, let's find the value of x by first equation.

$\sf \dashrightarrow 3x - y = -4$

$\sf \dashrightarrow 3x - 7 = -4$

$\sf \dashrightarrow 3x = -4 + 7$

$\sf \dashrightarrow 3x = 3$

$\sf \dashrightarrow x = \dfrac{3}{3}$

$\sf \dashrightarrow x = 1$

We know that, the values of x and y are 1 and 7 respectively.

Hence, the original fraction is $\sf \dfrac{1}{7}$.