Question

The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers. ​

Answer 1

Given question:-

The sum of the squares of two positive numbers is 233 and one number is 3 less than twice the other number. Find the numbers.

Solution:-

Given,

Sum of squares of two positive numbers = 233

One number + 3 = 2 × Other Number

• Let one number be 'x'
• Let another number be 'y'

According to the question,

~ Case | :-

x + 3 = 2y

x = 2y - 3 ......(1)

~Case || :-

${x}^{2} + {y}^{2} = 233.....(2)$

From equation, (1) and (2), we get,

${(2y - 3)}^{2} + {y}^{2} = 233$

${4y}^{2} - 9 - 12y + {y}^{2} = 233$

${5y}^{2} - 12y + 9 = 233$

${5y}^{2} - 12y + 9 - 233 = 0$

${5y}^{2} - 12y - 224 = 0$

${5y}^{2} - 40y + 28y- 224 = 0$

$5y(y - 8) + 28(y - 8) = 0$

$(y - 8)(5y + 28) = 0$

$y- 8 = 0 \\ y = 0 + 8 \\ y = 8$

Or

$5y+ 28 = 0 \\ 5y = 0 - 28$

$5y= - 28$

$y = \frac{ - 28}{5}$

Negative value is not acceptable. Hence, y = 8.

x = 2(8) -3

x = 16 -3

x = 13.