x+y =12 y =x^2/If(x,y) is a solution to the given system of equations, which of the following is a possible value of z? square A square B C 2 square D

To solve the system of equations, we can substitute the value of y from the second equation into the first equation:<br /><br />$x + \frac{x^2}{2} = 12$<br /><br />Multiplying both sides by 2 to eliminate the fraction, we get:<br /><br />$2x + x^2 = 24$<br /><br />Rearranging the terms, we have:<br /><br />$x^2 + 2x - 24 = 0$<br /><br />Factoring the quadratic equation, we get:<br /><br />$(x + 6)(x - 4) = 0$<br /><br />Setting each factor equal to zero, we find the possible values of x:<br /><br />$x + 6 = 0$ or $x - 4 = 0$<br /><br />$x = -6$ or $x = 4$<br /><br />Now, we can substitute these values of x back into the second equation to find the corresponding values of y:<br /><br />If $x = -6$, then $y = \frac{(-6)^2}{2} = 18$<br /><br />If $x = 4$, then $y = \frac{4^2}{2} = 8$<br /><br />Therefore, the possible values of z are 18 and 8.