Question

a)write a fraction with numerator x and denominator y
b)if its numerator increased by6 is 1/2
c)if its numerator increased by 7 is1/3​

Answer 1

Step-by-step explanation:

Let the fraction be

y

x

As per the given statements,

y+1

x−1

=

4

1

⇒4x−4=y+1

⇒4x−y=5 --- (1)

Also,

y−1

x+1

=

3

2

⇒3x+3=2y−2

⇒3x−2y=−5 --- (2)

Multi[plying eqn (1) by 2 and subtracting eqn (2), we get

2(4x−y)−(3x−2y)=2(5)−(−5)

⇒5x=15⇒x=3

Substituting this value of x in eqn (1),

4(3)−y=5

⇒y=7

∴x=3;y=7

So, the absolute difference between the numerator and denominator =7−3=4

Answer 2

Answer:

a) The fraction with numerator x and denominator y is $\frac{x}{y}$

b) The expression if the numerator of the fraction is  increased by 6 is $\frac{1}{2}$ = 2x-y = -12

c) The expression if the numerator of the fraction is  increased by 7 is $\frac{1}{3}$ = 3x-y = -21

Step-by-step explanation:

a) The fraction with numerator x and denominator y is $\frac{x}{y}$

b) If its numerator increased by 6 is $\frac{1}{2}$,

if the numerator is increased by 6, then the numerator becomes = x+6

Since when the numerator is increased by 6, the new fraction is  $\frac{1}{2}$, then we have

$\frac{x+6}{y} = \frac{1}{2}$

Cross multiplying we get

2(x+6) = y

2x+12 = y

2x-y = -12

The expression if the numerator of the fraction is  increased by 6 is $\frac{1}{2}$ is  2x-y = -12

c) If its numerator increased by 7 is $\frac{1}{3}$

Since when the numerator is increased by 7, the new fraction is  $\frac{1}{3}$, then we have

$\frac{x+7}{y} = \frac{1}{3}$

Cross multiplying we get

3(x+7) = y

3x+21 = y

3x-y = -21

The expression if the numerator of the fraction is  increased by 7 is $\frac{1}{3}$ is 3x-y = -21

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