Solve the system by the substitution or elimination method. $$\left( \begin{array} { l }
y& = &- x ^ { 2 } &+ &6 \
x &+& y &= &7
\end{array} \right)$$
A) $\left( \frac { 1 - i \sqrt { 3 } } { 2 } , \frac { 13 + i \sqrt { 3 } } { 2 } \right) , \left( \frac { 1 + i \sqrt { 3 } } { 2 } , \frac { 13 - i \sqrt { 3 } } { 2 } \right)$
B) $\left( \frac { 1 + i \sqrt { 3 } } { 2 } , \frac { 13 + i \sqrt { 3 } } { 2 } \right) , \left( \frac { 1 - i \sqrt { 3 } } { 2 } , \frac { 13 - i \sqrt { 3 } } { 2 } \right)$
C) $( 1 - i \sqrt { 3 } , 13 - i \sqrt { 3 } ) , ( 1 + i \sqrt { 3 } , 13 + i \sqrt { 3 } )$
D) $( 1 + i \sqrt { 3 } , 13 - i \sqrt { 3 } ) , ( 1 - i \sqrt { 3 } , 13 + i \sqrt { 3 } )$

Question

Solve the system by the substitution or elimination method. $$\left( \begin{array} { l } 
y& = &- x ^ { 2 } &+ &6 \
x &+& y &= &7
\end{array} \right)$$
A) $\left( \frac { 1 - i \sqrt { 3 } } { 2 } , \frac { 13 + i \sqrt { 3 } } { 2 } \right) , \left( \frac { 1 + i \sqrt { 3 } } { 2 } , \frac { 13 - i \sqrt { 3 } } { 2 } \right)$
B) $\left( \frac { 1 + i \sqrt { 3 } } { 2 } , \frac { 13 + i \sqrt { 3 } } { 2 } \right) , \left( \frac { 1 - i \sqrt { 3 } } { 2 } , \frac { 13 - i \sqrt { 3 } } { 2 } \right)$
C) $( 1 - i \sqrt { 3 } , 13 - i \sqrt { 3 } ) , ( 1 + i \sqrt { 3 } , 13 + i \sqrt { 3 } )$
D) $( 1 + i \sqrt { 3 } , 13 - i \sqrt { 3 } ) , ( 1 - i \sqrt { 3 } , 13 + i \sqrt { 3 } )$

Quizplus · Accepted Answer

Substituting $y = -x^2 + 6$ into $x + y = 7$, we get $x - x^2 + 6 = 7$, simplifying to $x^2 - x - 1 = 0$. Solving this quadratic equation using the quadratic formula gives $x = \frac{1 \pm i\sqrt{3}}{2}$. Substituting these values back into $y = -x^2 + 6$ gives the corresponding $y$ values, resulting in the complex solutions provided in option A.

Solve the System by the Substitution or Elimination Method (y=−x2+6x+y=7)\left( \begin{array} { l } y& = &- x ^ { 2 } &+ &6 \\x &+& y &= &7\end{array} \right)(yx​=+​−x2y​+=​67​)

Solve the System by the Substitution or Elimination Method $\left( \begin{array} { l } y& = &- x ^ { 2 } &+ &6 \\x &+& y &= &7\end{array} \right)$