Answer:
The points (-2, 0) and (3,5) are the ONLY solutions of the given sets of equations.
Explanation:
Here, the given equations are:
![y + 4 = x^(2) , y -x = 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bzvp60p93gxbxllxw459ido2l3ymveae6u.png)
Now checking for the given points:
(a) (-2, 0)
Here,
![y + 4 = 0 + 4 = 4 = (-2)^(2) = x^(2) \\y- x = 0 -(-2) = 2 = RHS](https://img.qammunity.org/2020/formulas/mathematics/middle-school/glt6509fonenhwu4cbjhhhgi6ozn2vbqux.png)
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,
![y + 4 = 0 + 4 = 4 = (2)^(2) = x^(2) \\y- x = 0 + (-2) = -2 \\eq 2(RHS)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/k3mo6e7bt0x76xppt26dbrcjhfd4ynppza.png)
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,
![y + 4 = 5 + 4 = 9 = (3)^(2) = x^(2) \\y- x = 5 -3 = 2 - (RHS)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hbq8choi58v5qm8u8mpv4fwxidi81uqgzo.png)
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.