Simultaneous equations: find the two intersects of x^2 + y^2 = 25 and y = 2x - 2

X^2 + y^2 = 25 (1)
y = 2x - 2 (2)

Subs (2) into (1)

x^2 + (2x - 2)^2 = 25
• Expand the eq like below

x^2 + (2x-2)(2x-2) = 25

x^2 + 4x^2 + 4x - 4x + 4 = 25
• Cancel out 4x-4x which is equals to zero

5x^2 = 25
• Divide both sides with 5

x^2 = 5
• To remove the square (^2) on the x, you need to square root both sides

x = 5 or x = -5

Subs x = 5 and x = -5 into (2)

y = 2x - 2
y = 2(5) - 2 = 10 - 2 = 8

y = 2x - 2
y = 2(-5) - 2 = -10 - 2 = -12

Therefore the intersection points are:

x=5, y= 8
x= -5, y= -12

40 because it’s just a little longer than 32 feet