Question

the sum of the numerator and denominator of a fraction is 4 more than twice the numerator. if the numerator and denominator each are increased by 3 the fraction becomes 2/3. find the original fraction​

Answer 1

Answer:

Let numerator =x and denominator=y .. Fraction = X

According to the first condition

⇒ x + y = 2x + 4 ⇒y=x+4

$According \: to \: the \: second \: condition</p><p></p><p>$

$\frac{x + 3}{y + 3} = \frac{2}{3}$

3x + 9 = 2y +6⇒ 3x - 2y =

-3....eq2

Substitute the value of y from eq1

to eq2

⇒ 3x − 2 (x + 4) = −3 ⇒ 3x - 2x + 8 =

-3⇒x=5

Put x = 5 in eq1

⇒y=5+4⇒y=9

Hence, fraction =

$\frac{5}{9}$

Answer 2

$\huge \rm {Answer:-}$

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$\sf\red{Consideration:-}$

$\mapsto \tt {Let,\: the\: Numerator\: be\: "x" }$

$\mapsto \tt {Let,\: the\: Denominator\: be\: "y" }$

$\mapsto \tt {And,\: the\: fraction\: be\: \frac{x}{y}}$

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★According to the conditions given in the Question:-

$\sf\purple{Condition\: 1:-}$

★The numerator and denominator of a fraction is 4 more than twice the numerator.

$\mapsto \tt{x+y=2x+4}$

$\mapsto \tt{y=2x-x+4}$

$\mapsto \tt{y=x+4}$ ------>Equation-1

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$\sf\pink{Condition\: 2:-}$

★The numerator and denominator each are increased by 3 the fraction becomes ${\frac{2}{3}}$.

$\large \mapsto \tt{\frac{x+3}{y+3}=\frac{2}{3}}$

★"As a result of cross multiplication",we get,

$\mapsto \tt {3x+9=2y+6}$

$\mapsto \tt{3x-2y=6-9}$

$\mapsto \tt{3x-2y=-3}$---->Equation-2

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$\sf\orange{Substitution:-}$

★"Substituting the value of 'y' from Equation-1 in Equation-2"

$\mapsto \tt {3x-2(x+4)=-3}$

$\mapsto \tt {3x-2x-8=-3}$

$\mapsto \tt {x=-3+8}$

$\large \tt {\fbox {x=5}}$

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★"As we got the value of x,it can substituted in either of the above given equations to get the value of y"

★-->>Substituting 'x value' in Equation-1

$\mapsto \tt {y=x+4}$

$\mapsto \tt {y=5+4}$

$\large \tt {\fbox {y=9}}$

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$\sf\blue{Thereupon,}$

★-->>The original fraction is found to be:

$\huge \mapsto \tt{\frac{x}{y}=\frac{5}{9}}$

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