Question: If three times the sum of two numbers is 42 and five times their difference is 20, find the numbers.

Solution:

Let the greater of the two numbers be 'x' and the other number be 'y'.

According to the question,

Condition I,

Three times the sum of two numbers is 42. 
or, 3(x+y) = 42
or, x + y = \frac{42}{3}
or, x + y = 14
or, x = 14 - y - (i)

Condition II,

Five times the difference of two numbers is 20.
or, 5(x -y) = 20
or, x - y = \frac{20}{5}
or, x - y = 4 - (ii)

Put value of x from equation i in equation ii, we get,
(14 - y) - y = 4
or, 14 - y - y = 4
or, 14 - 2y = 4
or, 14-4 -2y = 0
or, 10 = 2y
or, y = \frac{10}{2}
\therefore y = 5

Now,
Put value of y in equation i, we get,
or, x = 14 -5
\therefore x = 9

So, (x,y) = (9,5)

Hence, the required greater number is 9 and the required smaller number is 5 from the above mentioned conditions.

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