Question

6. If 15 is added to the numerator of a fraction, it becomes 3 and if 11 is aded to the denominator, it becomes
1/3
Find the fraction.

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Answer 1

Answer:

6/7

Step-by-step explanation:

let the fraction be x/y

x+15/y=3

x+15=3y

x=3y-15

take it as equation 1.

x/y+11=1/3

3x=y+11

substitute x

3(3y-15)=y+11

9y-45=y+11

9y-y=11+45

8y=56

y=56/8

y=7

substitute y in equation 1

x=3y-15

x=3×7-15

x=21-15

x=6

therefore the fraction is x/y=6/7

Answer 2

hope this helps  u

$\huge{\mathcal{\pink{A}\green{N}\red{S}\blue{W} {E}\green{R}}}$

let the fraction be $\frac{x}{y}$

case -(1)

If 15 is added to the numerator of a fraction, it becomes 3

$\frac{x\:+\:15}{y}$= 3

by cross multiplication

x+15= 3y

x-3y= -15

$\red{\boxed{x-3y= -15...(1)}}$

case -(2)

$\frac{x}{y\:+\:11}$= $\frac{1}{3}$

by cross multiplication

3x= y+11

3x-y= 11

$\red{\boxed{3x-y= 11...(2)}}$

multiply eq (1) by 3

x-3y= -15(*3)

3x-9y=-45

$\red{\boxed{3x-9y=-45....(3)}}$

subtract eq (3) and (2)

3x-9y= -45

3x-y= 11

(-) (+)   (-)

-----------

-8y= -56

y= $\frac{-56}{-8}$= 7

$\boxed{\red{y=7}}$

sub y = 7 in eq 2

3x-y=11

3x-7=11

3x = 11+7

3x = 18

x =  $\frac{18}{3}$=6

$\boxed{\red{x=6}}$

required fraction =  $\frac{6}{7}$