Question

The difference between the squares of two numbers is 11, three times the square of the first number increased by the square of the second number is 133 find the numbers

Answer 1

Let required numbers = x,y

The difference between the squres of two numbers is be

x^2-y^2 = 11 -------> (1)

Three times the square of the first number increased by the square of second number is be

3x^2+y^2 = 133 -------> (2)

To solve the varibles adding (1) and (2)

  x^2-y^2  = 11

3x^2+y^2 = 133

_______________

4x^2 = 144

Divide by 4 to each side.

4x^2/4 = 144/4

x^2 = 36

x = sqrt36

x = 6

Substitute the x value in (1)

6^2-y^2 = 11

-y^2 = 11-6^2

-y^2 = 11-36

-y^2 = -25

Cancell the signs to each side.

y =sqrt25

y = 5

Required numbers are x = 6, y = 5