Question

the sum of two number is 184 .one-third of one number exceed one-seventh of the other number by 8 .find the numbers.

Plz explain the answer also​

Answer 1

Step-by-step explanation:

Let the required numbers be x and y.

According to the first condition:

$x + y = 184\\</p><p>\implies x= 184-y.......(1)$

According to the second condition:

$\frac{1}{3} x = \frac{1}{7}y +8\\</p><p>\therefore \frac{x}{3} = \frac{y}{7} +8\\</p><p>\therefore\frac{x}{3} = \frac{y+56}{7}\\</p><p>\therefore 7x = 3(y+ 56)\\</p><p>\therefore 7x - 3y=168.......(2)$

From equations (1) & (2)

$7(184-y) - 3y = 168\\</p><p>\therefore 1288 - 7y - 3y = 168\\</p><p>\therefore 1288 - 168= 10y\\</p><p>\therefore 1120= 10y\\</p><p>\therefore y = \frac{1120}{10} \\</p><p>\huge\red{\boxed{ \therefore y = 112}} \\</p><p>\implies x = 184 - 112\\</p><p>\huge\blue {\boxed{\implies x = 72}}$