The difference between the sqaures of two numbers is 21. Twice the square of the first number increased by the square of the second number is 54.

Answer 1

Let’s take,

first no = x

2nd no = y

according to the question,

x²-y²=21

which gives x²=21+y² …….(1)

again last part of the question says,

2x²+y²=54 …… (2)

replacing x²=21+y² from eq. (1) into eq. (2) we have,

2(21+y²)+y² = 54

=> 42 +3y² = 54

=> 3y²= 12

=> y² = 4

=> y = (+/-) 2

replacing y = 2 into eqn (1)

x²=21+2²

=> x²=21+4

=> x²=25

=> x =(+/-) 5

so the two no are (+/-)5 and (+/-)2

Answer 2

x² - y² = 21

2x² + y² = 54.

Aftrer addition: 3x² = 75; x² = 25; x = ±5

y² 25 - 21 = 4, y = ±2.

We have 4 solutions: (5, 2); (- 5, 2); (5, - 2) and (- 5, - 2);