Question

The sum of two numbers is 184 one third of one number exceeds one seventh of the other other number by 8.

Answer 1

Let one number be "x"

and other be " y "

A/c , " The sum of two numbers is 184 "

⇒ x + y = 184 ... (1)

A/c , " one third of one number exceeds one seventh of the other other number by 8 "

$\implies \bold{\dfrac{1}{3}x=\dfrac{1}{7}y+8}\\\\\implies \bold{7x=3y+168}\\\\\implies \bold{7x-3y=168...(2)}$

Now solve (1) & (2) , i.e., 3(1) + (2).

$\implies \bold{3x+3y+7x-3y=552+168}\\\\\implies \bold{10x=720}\\\\\implies \bold{x=72}$

So other number , y = 184 - 72 = 112

Answer 2

Let the first number be x and the second number be y .

x + y = 184

y = 184 - x ____(1)

Also , $\frac{x}{3}$ = $\frac{y}{7}$ +8

$\implies \: \sf \: \frac{x}{3} = \frac{y + 56}{7}$

By cross - multiplication

$\sf \implies7x \: = 3(y + 56)$

$\sf \implies \: 7x = 3y + 168$

Using Y = 184 - x from equation 1

$\sf \implies \: 7x = 3(184 - x) + 168$

$\sf \implies \: 7x = 552 - 3x + 168$

$\sf \implies \: 7x + 3x = 552 + 168$

$\sf \implies \: 10x = 720$

$\sf \implies \: x = \frac{720}{10}$

$\implies \boxed{\sf x = 72}$

y = 184 - x

$\implies$y = 184 - 72

$\implies\boxed{ \sf \: y \: = 112}$

$\underline{ \rm\blue{verification : }}$

$\implies\frac{72}{3} = \frac{112}{7} + 8$

$\implies$ 24 = 16 +8

$\implies$ 24 = 24

Hence Verified