Question

a fraction becomes 9 by 11 if 2 is added to both the numerator and denominator if 3 is added to both the numerator and the denominator it becomes 5 by 6 find the fraction​

Answer 1

Answer:

Answer in explanation. Please mark me as brainliast. Hope it helps.

Step-by-step explanation:

Let the fraction be

y

x

Given that

y+2

x+2

=

11

9

⇒11x+22=9y+18

⇒11x−9y+4=0→i

⇒

y+3

x+3

=

6

5

⇒6x+18=5y+15

⇒6x−5y+3=0→ii

⇒i×5−ii×9

⇒55x−45y+20=0

⇒54x−45y+27=0

eq(i)−eq(ii)

⇒x−7=0

⇒x=7

⇒y=

9

11x+4

=

9

11(7)+4

=

9

81

=9

∴Fraction is

y

x

=

9

7

Please mark me as brainliast. Hope it helps.

Answer 2

Let the numerator of the fraction be x.

Let the numerator of the fraction be x.And the denominator of the fraction be y.

$\displaystyle{\sf\:The\:fraction\:=\:\dfrac{x}{y}}$

From the first condition,

$\displaystyle{\sf\:\dfrac{x\:+\:2}{y\:+\:2}\:=\:\dfrac{9}{11}}$

$\displaystyle{\implies\sf\:11\:\times\:(\:x\:+\:2\:)\:=\:9\:\times\:(\:y\:+\:2\:)}$

$\displaystyle{\implies\sf\:11x\:+\:22\:=\:9y\:+\:18}$

$\displaystyle{\implies\sf\:11x\:-\:9y\:=\:18\:-\:22}$

$\displaystyle{\implies\sf\:11x\:-\:9y\:=\:-\:4\:\:\:-\:-\:(\:1\:)}$

From the second condition,

$\displaystyle{\sf\:\dfrac{x\:+\:3}{y\:+\:3}\:=\:\dfrac{5}{6}}$

$\displaystyle{\implies\sf\:6\:\times\:(\:x\:+\:3\:)\:=\:5\:\times\:(\:y\:+\:3\:)}$

$\displaystyle{\implies\sf\:6x\:+\:18\:=\:5y\:+\:15}$

$\displaystyle{\implies\sf\:6x\:-\:5y\:=\:15\:-\:18}$

$\displaystyle{\implies\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:(\:2\:)}$

Multiplying equation ( 1 ) by 5 and equation ( 2 ) by 9, we get,

$\displaystyle{\sf\:11x\:-\:9y\:=\:-\:4\:\:\:-\:-\:(\:1\:)}$

$\displaystyle{\implies\sf\:5\:\times\:(\:11x\:-\:9y\:)\:=\:-\:4\:\times\:5}$

$\displaystyle{\implies\sf\:55x\:-\:45y\:=\:-\:20\:\:\:-\:-\:(\:3\:)}$

$\displaystyle{\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:(\:2\:)}$

$\displaystyle{\implies\sf\:9\:\times\:(\:6x\:-\:5y\:)\:=\:-\:3\:\times\:9}$

$\displaystyle{\implies\sf\:54x\:-\:45y\:=\:-\:27\:\:\:-\:-\:(\:4\:)}$

By subtracting equation ( 4 ) from equation ( 3 ), we get,

$\displaystyle{\sf\:55x\:-\:45y\:-\:(\:54x\:-\:45y\:)\:=\:-\:20\:-\:(\:-\:27\:)}$

$\displaystyle{\implies\sf\:55x\:-\:\cancel{45y}\:-\:54x\:+\:\cancel{45y}\:=\:-\:20\:+\:27}$

$\displaystyle{\implies\sf\:55x\:-\:54x\:=\:7}$

$\displaystyle{\implies\boxed{\red{\sf\:x\:=\:7}}}$

By substituting x = 7 in equation ( 2 ), we get,

$\displaystyle{\sf\:6x\:-\:5y\:=\:-\:3\:\:\:-\:-\:-\:(\:2\:)}$

$\displaystyle{\implies\sf\:6\:\times\:7\:-\:5y\:=\:-\:3}$

$\displaystyle{\implies\sf\:42\:-\:5y\:=\:-\:3}$

$\displaystyle{\implies\sf\:-\:5y\:=\:-\:3\:-\:42}$

$\displaystyle{\implies\sf\:\cancel{-}\:5y\:=\:\cancel{-}\:45}$

$\displaystyle{\implies\sf\:5y\:=\:45}$

$\displaystyle{\implies\sf\:y\:=\:\cancel{\dfrac{45}{5}}}$

$\displaystyle{\implies\boxed{\purple{\sf\:y\:=\:9}}}$

$\displaystyle{\therefore\:\underline{\boxed{\blue{\sf\:The\:fraction\:=\:\dfrac{7}{9}}}}}$

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