use the method of substitution to solve the system.?

Question

solve the system.
xsqaured +ysquared =16
x+y=1
what are the solutions for x and y

Answer 1

Let x=1-y

Substitute into x^2 + y^2 =16, you get y = 1/2 ((sq.root 31)+1)

Plug into x+y=1, you get x=(1/2) (1 - (sq.root31))

Answer 2

x^2 + y^2 = 16 ----> equation number 1

x + y = 1 ----> equation number 2

From equation number 2:
x + y = 1

So,
x = 1 - y ----> equation number 3

Substitute equation number 3 into equation number 1:
(1 - y)^2 + y^2 = 16

Solve the above equation as usual then substitute the value of y into equation number 3 to get the value of x.

Good luck!

Answer 3

Values of y:
x^2 + y^2 = 16
x + y = 4
y = 4 - x

x + y = 1
y = 1 - x

Value of x:
4 - x = 1 - x
No definite answer.

Answer 4

y = 1 - x
x² + (1 - x)² = 16
x² + 1 - 2x + x² = 16
2x² - 2x - 15 = 0
x = [ 2 ± √(4 + 120)] / 4
x = [ 2 ± √(124) ] / 4
x = [ 2 ± 2√(31) ] / 4
x = (1/2) [ 1 ± √(31) ]
y = 1 - (1/2) [ 1 ± √(31) ]

Answer 5

x=1-y

so

x^2 + (1-y)^2
x^2 + 1 - 2y + y^2 = 16
y^2 - 2y + __ + x^2 +1 = 16 + __
y^2 - 2y + 1 + x^2 + 1 = 16 + 1
(y-1)(y-1)^2 + (x)^2 +1 = 17

x has no value

there is no answer
i think, i might have fu |{ E d up O.o

eh, i tried

Answer 6

it cannot be.