Answer:
The first multiple of 7 = 14 or -21
The second multiple of 7 = 21 and -14
Step-by-step explanation:
Let
The first multiple of 7 = 7x
The second multiple of 7 = 7(x+1)
Square of the first multiple = (7x)^2
Square of second multiple = {7(x+1)}^2
Sum
(7x)^2 + 7(x+1)}^2 = 637
49x^2 + 49(x+1)^2 = 637
Divide through by 49
x^2 + (x+1) = 13
Open bracket
x^2 + x^2 + 1 + 2x = 13
Subtract 13 from both sides
x^2 + x^2 + 1 + 2x - 13 = 13 - 13
x^2 + x^2 + 1 + 2x - 13 = 0
2x^2 + 2x - 12 = 0
Divide through by 2
x^2 + x - 6 = 0
Solve the quadratic equation using factorization method
x^2 + 3x - 2x - 6 = 0
x(x+3) - 2(x+3) = 0
(x - 2) (x+3) = 0
x - 2 = 0
x = 2
x+3 = 0
x = -3
Substitute x = 2 or -3 into 7x
When x= 2
7(2) = 14
When x = -3
7(-3) = -21
The first multiple of 7 = 14 or -21
Substitute x = 2 or -3 into 7(x+1)
When x= 2
7(2+1) = 7(3) = 21
When x = -3
7(-3+1) = 7(-2) = -14
The second multiple of 7 = 21 and -14
