Solve the system of equations algebraically.
$$\left\{ \begin{array} { r }
x y + 289 = 0 \
4 x - 25 y - 340 = 0
\end{array} \right.$$
A) $\left( 85 , - \frac { 17 } { 5 } \right)$
B) $\left( - 85 , \frac { 17 } { 5 } \right) , \left( 85 , - \frac { 17 } { 5 } \right)$
C) $\left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
D) $\left( - \frac { 85 } { 2 } , \frac { 34 } { 5 } \right) , \left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
E) no solution

Question

Solve the system of equations algebraically.
$$\left\{ \begin{array} { r } 
x y + 289 = 0 \
4 x - 25 y - 340 = 0
\end{array} \right.$$
A) $\left( 85 , - \frac { 17 } { 5 } \right)$
B) $\left( - 85 , \frac { 17 } { 5 } \right) , \left( 85 , - \frac { 17 } { 5 } \right)$
C) $\left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
D) $\left( - \frac { 85 } { 2 } , \frac { 34 } { 5 } \right) , \left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
E) no solution

Quizplus · Accepted Answer

The Answer of Solve the system of equations algebraically.
$$\left\{ \begin{array} { r } 
x y + 289 = 0 \
4 x - 25 y - 340 = 0
\end{array} \right.$$
A) $\left( 85 , - \frac { 17 } { 5 } \right)$
B) $\left( - 85 , \frac { 17 } { 5 } \right) , \left( 85 , - \frac { 17 } { 5 } \right)$
C) $\left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
D) $\left( - \frac { 85 } { 2 } , \frac { 34 } { 5 } \right) , \left( \frac { 85 } { 2 } , - \frac { 34 } { 5 } \right)$
E) no solution

Solve the System of Equations Algebraically {xy+289=04x−25y−340=0\left\{ \begin{array} { r } x y + 289 = 0 \\4 x - 25 y - 340 = 0\end{array} \right.{xy+289=04x−25y−340=0​

Solve the System of Equations Algebraically $\left\{ \begin{array} { r } x y + 289 = 0 \\4 x - 25 y - 340 = 0\end{array} \right.$