Answer:
The points (-2, 0) and (3,5) are the ONLY solutions of the given sets of equations.
Step-by-step explanation:
Here, the given equations are: [tex]y + 4 = x^{2} , y -x = 2[/tex]
Now checking for the given points:
(a) (-2, 0)
Here, [tex]y + 4 = 0 + 4 = 4 = (-2)^{2} = x^{2} \\y- x = 0 -(-2) = 2 = RHS[/tex]
Hence, (-2, 0) is the solution of the given equations.
b) Checking for (2,0), as (-2, 0) is a solution as shown above
Here, [tex]y + 4 = 0 + 4 = 4 = (2)^{2} = x^{2} \\y- x = 0 + (-2) = -2 \neq 2(RHS)[/tex]
Hence, (2, 0) is NOT the solution of the given equations.
c) Checking for (3,5), as (-2, 0) is a solution as shown above
Here, [tex]y + 4 = 5 + 4 = 9 = (3)^{2} = x^{2} \\y- x = 5 -3 = 2 - (RHS)[/tex]
Hence, (3,5), is the solution of the given equations.
Hence, the points (-2, 0) and (3,5) are the ONLY solutions of the given sets of equations.
