What is the solution set to. the following system?{x+y=4, x^2+y^2=16}

Question

Thaddeus Bray

Mathematics

26 February, 11:53

What is the solution set to. the following system?{x+y=4, x^2+y^2=16}

+2

Answers (1)

Know the Answer?

Brian Sexton · Answer 1

x = 3.47 or x = - 4.47, y = 0.53 or y = 8.47

Step-by-step explanation:

quadratic equation:

x^2 + y^2 = 16

(-b± √b^2 - 4*a*c) / 2

(-1 ± √ 1-4*1*16) / 2

(-1 ± 3√7) / 2

(-1 ± 7.94) / 2

(-1 + 7.94) / 2 or (-1-7.94) / 2

x = 3.47 or - 4.47

therefore:

x+y = 4

3.47 + y = 4 or - 4.47 + y = 4

y = 4-3.47 or y = 4 + 4.47

y = 0.53 or y = 8.47