Question

A fraction reduces to when ¼ when 2 is subtracted from the
numerator and 3 is added to the denominator. But it reduce
to 2/3 if 6 is added to the numerator and denominator is
multiplied by 3. Find the fraction.

plezzzzzzzzzzzzzzzzz giv me fast answer with expanetion

Answer 1

SOLUTION :-

Let,

Numerator = x

Denominator = y

According to the first condition,

$\longrightarrow\sf \dfrac{x-2}{y+3}=\dfrac{1}{4}\\\\\longrightarrow 4(x-2)= y+3\\\\\longrightarrow 4x-8 = y+3\\\\\longrightarrow 4x-y = 11 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .........................(1)$

According to the second condition,

$\longrightarrow\sf \dfrac{x+6}{3\times y }=\dfrac{2}{3} \\\\\longrightarrow 3(x+6) = 2\times 3y \\\\\longrightarrow 3x+18 = 6y \\\\\longrightarrow 3x-6y = - 18 \\\\\longrightarrow x-2y = -6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .........................(2)$

Multiply equation (2) by 4,

$\longrightarrow\sf 4x-8y= -24 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .........................(3)$

Equation (1) - Equation (3),

$\longrightarrow \sf 7y = 35\\\\\longrightarrow\bf y = 5$

Substitute the value of y in equation (2),

$\longrightarrow\sf x- 2\times 5 = - 6 \\\\\longrightarrow x-10 = -6 \\\\\longrightarrow x = -6+10\\\\\longrightarrow\bf x = 4$

$\bf FRACTION = \dfrac{4}{5}$

Answer 2

AnsWeR :-

A fraction reduces to when ¼ when 2 is subtracted from the numerator and 3 is added to the denominator. But it reduce to 2/3 if 6 is added to the numerator and denominator is multiplied by 3. Find the fraction.

The Given question is this From the Mentioned Question

GiveN :-

• A Fraction reduces to ¼ when 2is substracted from the Numerator and 3 is added to the Denominator.

But In the Question it also States That, the second Condition of the Question.

• But it is reduced to ⅔ when 6 is added to the Numerator and Denominator is Multiplied by 3.

To FinD :-

• To Find the required Fraction.

SoluTioN :-

We Can Consider given two Mentioned Conditions step by step

$\boxed{ \bold{ \pink{★Case \: 1}{} }{} }{}$

Let,

Numerator = x

Denominator = y

$\frac{x - 2}{y + 3} = \frac{1}{4} \\ \\ 4(x - 2) = y + 3 \\ \\ 4x - 8 = y + 3 \\ \\ 4x - y = 11 - - - (1)$

$\boxed{ \bold{ \pink{★Case \: 2}{} }{} }{}$

In This case we discuss the ll Condition mentioned in the Question.

$\frac{x + 6}{3 \times y} = \frac{2}{3} \\ \\ 3(x + 6) = 2 \times 3y \\ \\ 3x + 18 = 6y \\ \\ 3x - 6y = - 18 \\ \\ = > x - 2y = - 6 - - - (2)$

Now In Equation (1) - Equation (3)

$7y = 35 \\ \\ y = 5$

Substitute The Value of 'y' in Equation (2) we get,

$x - 2 \times 5 = - 6 \\ \\ x - 10 = - 6 \\ \\ x = - 6 + 10 \\ \\ x = 4$

Hence We Conclude That,

$\boxed{ \bold{ \red{★FrActTiOn = \frac{4}{5} }{} }{} }{}$

Hence Prooved that the required Fraction = 4/5.