Answer :
Let smaller number= x & larger number= y
According to given condition,
y^2 - x^2 = 88 ..........(1)
y = 2x - 5 ...............(2)
Put value of y frm equation 2 in eq 1
(2x-5)^2 - x^2 = 88
⇒ 4x^2 - 20x +25 -x^2 -88 = 0
⇒3x^2 -20x -63 = 0
⇒3x^2 -27x+7x-63 =0
⇒3x(x-9) +7(x-9) = 0
⇒(3x+7)(x-9) =0
⇒x=9, x= -7/3 (as no. cant b -ve, so x=9)
Hence, smaller no = 9
& larger no = 2x-5 = 2*9-5 = 18-5 =13
According to given condition,
y^2 - x^2 = 88 ..........(1)
y = 2x - 5 ...............(2)
Put value of y frm equation 2 in eq 1
(2x-5)^2 - x^2 = 88
⇒ 4x^2 - 20x +25 -x^2 -88 = 0
⇒3x^2 -20x -63 = 0
⇒3x^2 -27x+7x-63 =0
⇒3x(x-9) +7(x-9) = 0
⇒(3x+7)(x-9) =0
⇒x=9, x= -7/3 (as no. cant b -ve, so x=9)
Hence, smaller no = 9
& larger no = 2x-5 = 2*9-5 = 18-5 =13
Let the numbers be x (smaller) and y (larger).
According to the problem,
y² - x² = 88 ... (i)
y = 2x - 5 ... (ii)
Substituting the value of y from the equation (ii) in equation (i), we get,
(2x - 5)² - x² = 88
⇒ 4x² - 20x + 25 - x² - 88 = 0
⇒ 3x² - 20x - 63 = 0
⇒ 3x² - 27x + 7x - 63 = 0
⇒ 3x (x - 9) + 7 (x - 9) = 0
⇒ (3x + 7) (x - 9) = 0
∴ Either x = 9 or x = - 7/3
But x cannot be negative.
So, x = 9
And, y = 2x - 5 = 2 × 9 - 5 = 18 - 5 = 13
According to the problem,
y² - x² = 88 ... (i)
y = 2x - 5 ... (ii)
Substituting the value of y from the equation (ii) in equation (i), we get,
(2x - 5)² - x² = 88
⇒ 4x² - 20x + 25 - x² - 88 = 0
⇒ 3x² - 20x - 63 = 0
⇒ 3x² - 27x + 7x - 63 = 0
⇒ 3x (x - 9) + 7 (x - 9) = 0
⇒ (3x + 7) (x - 9) = 0
∴ Either x = 9 or x = - 7/3
But x cannot be negative.
So, x = 9
And, y = 2x - 5 = 2 × 9 - 5 = 18 - 5 = 13