Question 430547: Please help me solve this problem by Substitution Method 
 I can't figure out the solution sets.
Thank you
Found 2 solutions by sudhanshu_kmr, Edwin McCravy:
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
x-y = 1 , so x= y+1
substitute the value of x in
x^2 -xy -y^2 =-5
=> (1+y)^2 -(1+y)y - y^2 = -5
=> 1 + y^2 +2y -y - y^2 -y^2 = -5
=> y^2 -y -6 = 0
=> (y+2)(y-3) =0
y = -2 or y =3
when y = -2, x= -1
when y = 3 , x = 4
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
x - y = 1
x² - xy - y² = -5
The other tutor solved the first equation for x.
Just to be different I'll solve it for y:
x - y = 1
-y = 1 - x
y = -1 + x
y = x - 1
Substitute (x - 1) for y in the second equation,
but first factor -y out of the second and third terms
on the left:
x² - xy - y² = -5
x² - y(x + y) = -5
x² - (x - 1)[x + (x - 1)] = -5
x² - (x - 1)(x + x - 1) = -5
x² - (x - 1)(2x - 1) = -5
x² - (2x² - 3x + 1) = -5
x² - 2x² + 3x - 1 = -5
-x² + 3x + 4 = 0
x² - 3x - 4 = 0
(x - 4)(x + 1) = 0
x - 4 = 0; x + 1 = 0
x = 4 x = -1
To find the y-value for x = 4, substitute 4 for x in
y = x - 1
y = 4 - 1
y = 3
So one solution is (x,y) = (4,3)
To find the y-value for x = -1, substitute -1 for x in
y = x - 1
y = -1 - 1
y = -2
So the other solution is (x,y) = (-1,-2)
Edwin
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