Solve the System

Question

Solve the system of equations using any method.

x^2+y=13

x^2+y^2=25

Answer 1

let z = x²

z+y = 13

z+y²=25

Subtract the 1st equation from the second

y²-y = 12 -->(y-4)(y+3) = 0 --> y = 4,-3

x² + 4 = 13 --> x = ±√9 = ±3

x²-3 = 13 -> x = ±√16 = ±4

x = -3,3,-4,4

y = 4, -3

Answer 2

that's a circle and a downward opening parabola intersecting in potentially 4 points

subtract 1st from 2nd to get

y^2 -y = 12

y^2 -y - 12 = 0

factor

(y-4)(y+3) = 0

y = -3 or 4, plug those 2 values into either original equation to get 2 x values

x^2 + 3 = 13

x^2 =10

x = sqr(10)

the point (sqr10,3) or (-sqr10,3)

or

x^2 +4 = 13

x^2 = 9

x = + or - 3

giving he points (3,4) or ( -3,4)

4 intersections

(3,4), (-3,4), (sqr10,3) and (-sqr10.3,3)