Solve the system of equations using substitution.
-$$\left\{ \begin{array} { r }
x y = 20 \
x + y = 9
\end{array} \right.$$
A) $x = 20 , y = 1 ; x = 1 , y = 20$
B) $x = 10 , y = 2 ; x = 2 , y = 10$
C) $x = 6 , y = 3 ; x = 3 , y = 6$
D) $x = 5 , y = 4 ; x = 4 , y = 5$

Question

Solve the system of equations using substitution.
-$$\left\{ \begin{array} { r } 
x y = 20 \
x + y = 9
\end{array} \right.$$
A) $x = 20 , y = 1 ; x = 1 , y = 20$
B) $x = 10 , y = 2 ; x = 2 , y = 10$
C) $x = 6 , y = 3 ; x = 3 , y = 6$
D) $x = 5 , y = 4 ; x = 4 , y = 5$

Quizplus · Accepted Answer

From the second equation, $y = 9 - x$. Substituting this into the first equation gives $x(9 - x) = 20$, which simplifies to $x^2 - 9x + 20 = 0$. Factoring this quadratic equation gives $(x - 5)(x - 4) = 0$, leading to $x = 5$ or $x = 4$. Substituting these values back into $y = 9 - x$ gives the pairs $(5, 4)$ and $(4, 5)$.

Solve the System of Equations Using Substitution {xy=20x+y=9\left\{ \begin{array} { r } x y = 20 \\x + y = 9\end{array} \right.{xy=20x+y=9​

Solve the System of Equations Using Substitution $\left\{ \begin{array} { r } x y = 20 \\x + y = 9\end{array} \right.$