Solve for x and y?

Question

x^2 + xy - y^2 = -5

x + y = 1

Thanks for the help!

Answer 1

x^2+ x(1-x)- (1-x)^2 = -5 From eq 2, y= 1-x
x - (1-x)^2 = -5 Partial multout & combine
(1-x)^2 - x - 5 = 0 Rearrange, mult by -1
1-2x+x^2-x-5=0 Expand
x^2-3x-4=0 Combine
(x-4)(x+1)=0 Factor
x= 4 or -1
y = -3 or +2

Answer 2

x^2 + xy - y^2 = -5

x^2 - y^2 + xy + 5 = 0

is this a system?

If so, then solve for x or y in the second equation

x + y = 1

x = -y + 1

plug this into the first equation to solve for y.

(-y + 1)^2 + (-y + 1)(y) - y^2 + 5 = 0

y^2 + 1 + -y^2 + y - y^2 + 5 = 0

combine like terms
-y^2 + y + 6 = 0

factor and solve for y
(-y - 2)(y - 3)
y = 3
y = -2

Plug those values for y back into the second original equation that was solved for x:
x = -y + 1

x = -3 + 1 and x = 2 + 1
x = -2 x = 3

Answer 3

Set x + y =1 to x = 1 - y

thus plug in for x in the equation... x^2 + xy - y^2 = -5

(1 - y)^2 + (1 - y) y - y^2 = -5
(1 - y)(1 - y) + y - y^2 - y^2 = - 5
(1 - 2y + y^2) + y - y^2 - y^2 = - 5
y - y^2 = -6

wow i just confused my self.

The answer could be No Solution, Inconsistent. Or i can be wrong sorry! :[

Answer 4

x + y = 1
x = 1 - y

x^2 + xy - y^2 = -5
x ( x + y) - y^2 = -5
(1-y) (1-y + y) - y^2 = -5

y^2 + y -6 = 0
(y - 2) ( y + 3) = 0
y = 2, -3
x = -1, 4

Answer 5

x=1-y
substitute (1-y) for x in the first equation...

(1-y)^2+(1-y)y-y^2=-5
1-2y+y^2+y-y^2-y^2=-5

now subtract... and you get
1-y-y^2=-5

move every term to the right side...
0=y^2+y-6

factor...
0=(y+3)(y-2)
y=2 or -3

plug in each of the y value to the second equation to get x...
when y=2, x+2=1 so x=-1

when y=-3, x-3=1 so x=4

answer: x=4, y=-3 or x=-1, y=2

ordered pair: (4,-3) and (-1,2)

Answer 6

x= 4
y= -3

I guess i'm too late. The answer above is the best answer.

Answer 7

x = 1 - y

now substitute

Answer 8

Sorry teacher! oh i mean Sorry..............bumphead