Question

The numerator of a fraction is 6 less than the denominator. If 1 is added to both numerator and denominator, the fraction becomes 1/2. The equation for this is ......

Answer 1

Answer:

Let the denominator = x

Numerator = x - 6

Original fraction = ( x - 6 ) / x ---(a)

Follow the question as per as

( x - 6 + 3 ) / x = 2/3

( x - 3 ) / x = 2 / 3

3( x - 3 ) = 2x

3x - 9 = 2x

3x - 2x = 9

x = 9

ACCORDING TO THE QUESTION;

Original/Required fraction = ( x - 6 ) / x

= ( 9 - 6 ) / 9

= 3 / 9

Step-by-step explanation:

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Answer 2

ANSWER:

The two required equations are:

x = y - 6

$\sf \dfrac{x + 1}{y+1} =\dfrac{1}{2}$

GIVEN:

• The numerator of a fraction is 6 less than the denominator.

• If 1 is added to both numerator and denominator, the fraction becomes 1/2.

EXPLANATION:

Let the numerator be x and the denominator be y.

x = y - 6

$\sf \dfrac{x + 1}{y+1} =\dfrac{1}{2}$

These are the equations used to solve this problem.

SOLUTION:

$\sf \dfrac{x + 1}{y+1} =\dfrac{1}{2}$

Substitute x = y - 6

$\sf \dfrac{y - 6 + 1}{y+1} =\dfrac{1}{2}$

$\sf \dfrac{y - 5}{y+1} =\dfrac{1}{2}$

$\sf 2(y - 5) = 1(y+1)$

$\sf 2y - 10= y+1$

$\sf y=11$

Substitute y = 11 in x = y - 6

x = 11 - 6

x = 5

Numerator = 5, Denominator = 11.

Hence the fraction = 5/11.

VERIFICATION:

$\sf \dfrac{x + 1}{y+1} =\dfrac{1}{2}$

Substitute x = 5 and y = 11

$\sf \dfrac{5 + 1}{11+1} =\dfrac{1}{2}$

$\sf \dfrac{6}{12} =\dfrac{1}{2}$

$\sf \dfrac{1}{2} =\dfrac{1}{2}$

HENCE VERIFIED.