Question

in a fraction twice the numerator is 2 more than the denominator if 3 is added to the numerator and the denominator the new fraction is 2/3 find the original fraction​

Answer 1

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$\large{\underline{\bf{\pink{Answer:-}}}}$

✰ Fraction = 7/12

$\large{\underline{\bf{\blue{Explanation:-}}}}$

$\large{\underline{\bf{\green{Given:-}}}}$

✰ numerator is 2 more than then twice it's denominator.

✰ if 3 is added to both numerator and denominator the fraction becomes = 2/3.

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the original fraction.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let the numerator be x

so , the denominator = 2x -2

$\blue {\underbrace{According\:to\: question}}$

If we add 3 to both numerator and denominator

the fraction becomes 2/3.

$:\implies\bf\frac{x+3}{2x-2+3}=\frac{2}{3}$

$:\implies$ 3( x + 3) = 2( 2x + 1)

$:\implies$ 3x + 9 = 4x + 2

$:\implies$ 9 - 2 = 4x - 3x

$:\implies$ x = 7

So the numerator (x) = 7

And the denominator = 2x - 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 14 -2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 12

So the fraction = 7/12

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

Fraction is $\bold{\frac{7}{12}}$

$\bold\green{\underline{\underline{Solution}}}$

Let the numenator be x and denominator be y.

According to the first condition

2x=y+2

2x-y=2...(1)

According to the second condition

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

$\bold{3(x+3)=2(y+3)}$

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

$\bold{3x-2y=-3...(2)}$

Multiply equation (1) by 2

4x-2y=4...(3)

Subtract equation (2) from equation (3), we get

4x-2y=4

-

3x-2y=-3

x=7

Substituting x=7 in equation (1), we get

2(7)-y=2

14-y=2

-y=4-14

-2y=-12

$\bold{y=\frac{-12}{-1}}$

y=12

Fraction is $\bold{\frac{x}{y}}$

i.e $\bold{\frac{7}{12}}$