Solve the system of equations using substitution.
-$$\left\{ \begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 4 \
x + y = 2
\end{array} \right.$$
A) $x = 0 , y = 0 ; x = 2 , y = - 2$
B) $x = 0 , y = - 2 ; x = - 2 , y = 0$
C) $x = 0 , y = 2 ; x = 2 , y = 0$
D) $x = 2 , y = - 2 ; x = - 2 , y = - 2$

Question

Solve the system of equations using substitution.
-$$\left\{ \begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 4 \
x + y = 2
\end{array} \right.$$
A) $x = 0 , y = 0 ; x = 2 , y = - 2$
B) $x = 0 , y = - 2 ; x = - 2 , y = 0$
C) $x = 0 , y = 2 ; x = 2 , y = 0$
D) $x = 2 , y = - 2 ; x = - 2 , y = - 2$

Quizplus · Accepted Answer

From the second equation, $y = 2 - x$. Substituting this into the first equation gives $x^2 + (2-x)^2 = 4$, which simplifies to $2x^2 - 4x + 4 = 4$, and further to $2x^2 - 4x = 0$. Factoring out $2x$ gives $2x(x - 2) = 0$, leading to $x = 0$ or $x = 2$. Substituting these values back into $y = 2 - x$ gives the pairs $(0, 2)$ and $(2, 0)$.

Solve the System of Equations Using Substitution {x2+y2=4x+y=2\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 4 \\x + y = 2\end{array} \right.{x2+y2=4x+y=2​

Solve the System of Equations Using Substitution $\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 4 \\x + y = 2\end{array} \right.$