Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 - y^2 = 56.
x + y = 10 Subtract y from both sides of this equation to get x = 10 - y
Now since x = 10 - y we can substitute 10 - y in for x in the second equation.
x2 - y2 = 56
(10 - y)2 - y2 = 56
(10 - y)(10 - y) - y2 = 56
100 - 20y + y2 - y2 = 56
100 - 20y = 56
-20y = -44
y = 2.2
Using this value of y and the fact that x = 10 - y we can find the value of x
x = 10 - y = 10 - 2.2 = 7.8
So there is only one ordered pair of real numbers that satisfies the system of equations, and it is (7.8, 2.2)
x + y = 10 Subtract y from both sides of this equation to get x = 10 - y
Now since x = 10 - y we can substitute 10 - y in for x in the second equation.
x2 - y2 = 56
(10 - y)2 - y2 = 56
(10 - y)(10 - y) - y2 = 56
100 - 20y + y2 - y2 = 56
100 - 20y = 56
-20y = -44
y = 2.2
Using this value of y and the fact that x = 10 - y we can find the value of x
x = 10 - y = 10 - 2.2 = 7.8
So there is only one ordered pair of real numbers that satisfies the system of equations, and it is (7.8, 2.2)