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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