Help me Please
Solve the Systems of equations
{x+y=5}
{x2+y2=13}
x + y = 5 Subtract y from both sides of this equation.
x = 5 - y Use this value of x in the second equation.
x2 + y2 = 13
(5 - y)2 + y2 = 13
(5 - y)(5 - y) + y2 = 13
25 - 10y + y2 + y2 = 13
25 - 10y + 2y2 = 13 Subtract 13 from both sides of the equation.
12 - 10y + 2y2 = 0 Rearrange.
2y2 - 10y + 12 = 0 Divide through by 2 .
y2 - 5y + 6 = 0 Factor left side.
(y - 3)(y - 2) = 0 Set each factor equal to zero.
y - 3 = 0 or y - 2 = 0
y = 3 or y = 2
Use both of these values of y to find x .
When y = 3 , x = 5 - y = 5 - 3 = 2
When y = 2 , x = 5 - y = 5 - 2 = 3
So...
One solution is x = 2 and y = 3
Another solution is x = 3 and y = 2
The solution set is { (2, 3), (3, 2) }
x + y = 5 Subtract y from both sides of this equation.
x = 5 - y Use this value of x in the second equation.
x2 + y2 = 13
(5 - y)2 + y2 = 13
(5 - y)(5 - y) + y2 = 13
25 - 10y + y2 + y2 = 13
25 - 10y + 2y2 = 13 Subtract 13 from both sides of the equation.
12 - 10y + 2y2 = 0 Rearrange.
2y2 - 10y + 12 = 0 Divide through by 2 .
y2 - 5y + 6 = 0 Factor left side.
(y - 3)(y - 2) = 0 Set each factor equal to zero.
y - 3 = 0 or y - 2 = 0
y = 3 or y = 2
Use both of these values of y to find x .
When y = 3 , x = 5 - y = 5 - 3 = 2
When y = 2 , x = 5 - y = 5 - 2 = 3
So...
One solution is x = 2 and y = 3
Another solution is x = 3 and y = 2
The solution set is { (2, 3), (3, 2) }