Question

If 1 is added to both numerator and denominator of a fraction it becomes equal to 7/8 however 1 is subtracted from both numerator and denominator of the same fraction it becomes equal to 6 /7 find the fraction let the required fraction be

Answer 1

ANSWER = 13/15

$\sf \: Let \: the \: fraction \: be \: \: \frac{x}{y}$

$\sf \frac{x + 1}{y + 1} = \frac{7}{8}$

$\sf \: 8(x + 1) = 7(y + 1)$

$\sf8x + 8 = 7y + 7$

$8x - 7y = - 1 - - - - (1)$

$\rule{10cm}{0.02cm}$

$\sf \: \frac{x - 1}{y - 1} = \frac{6}{7}$

$\sf \: 7(x - 1) = 6(y - 1)$

$\sf7x - 7 = 6y - 6$

$\sf7x - 6y = 1 - - - - - - (2)$

$\rule{10cm}{0.02cm}$

$8x - 7y = - 1 - - - - - - - (1)$

$7x - 6y = 1 - - - - - - - (2)$

$\rule{10cm}{0.02cm}$

$\sf(1) \times 7 \implies \: 56x - 49y = - 7 - - (3)$

[

$\sf(2) \times 8\implies56x - 48y = 8 - - (4)$

$\sf \: (3) - ( - (4)) \implies \: \: - y = - 15$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y \: \: \: \: \: = 15$

$\rule{10cm}{0.02cm}$

substitute the value of y in (1)

$\implies \: 8x - 7( 15)= - 1$

$\implies \: 8x - 105= - 1$

$\implies \: 8x = 104$

$\implies \: x \: = \frac{ 104}{8}$

$\implies \: x \: = 13$

$\rule{10cm}{0.02cm}$

$fraction \: = \frac{13}{15}$

Answer 2

Answer:13/15

Step-by-step explanation: