Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { r }
20 x - y = 1 \
x y = 1
\end{array} \right.$
A) $\left( \frac { 1 } { 4 } , 4 \right) , \left( - \frac { 1 } { 5 } , - 5 \right)$
B) $\left( 4 , \frac { 1 } { 4 } \right) , \left( - 5 , - \frac { 1 } { 5 } \right)$
C) $\varnothing$
D) $( - 4,4 ) , ( 5 , - 5 )$

Question

Solve the nonlinear system of equations for real solutions.
-$\left\{ \begin{array} { r } 
20 x - y = 1 \
x y = 1
\end{array} \right.$
A) $\left( \frac { 1 } { 4 } , 4 \right) , \left( - \frac { 1 } { 5 } , - 5 \right)$
B) $\left( 4 , \frac { 1 } { 4 } \right) , \left( - 5 , - \frac { 1 } { 5 } \right)$
C) $\varnothing$
D) $( - 4,4 ) , ( 5 , - 5 )$

Quizplus · Accepted Answer

From the second equation $xy = 1$, we can express $y$ as $y = \frac{1}{x}$. Substituting this into the first equation gives $20x - \frac{1}{x} = 1$. Multiplying through by $x$ to clear the denominator gives $20x^2 - 1 = x$, or $20x^2 - x - 1 = 0$. Solving this quadratic equation for $x$ yields $x = \frac{1}{4}$ and $x = -\frac{1}{5}$. Substituting these values back into $y = \frac{1}{x}$ gives the corresponding $y$ values as $4$ and $-5$, respectively. Therefore, the solutions are $\left( \frac{1}{4}, 4 \right)$ and $\left( -\frac{1}{5}, -5 \right)$.

Solve the Nonlinear System of Equations for Real Solutions {20x−y=1xy=1\left\{ \begin{array} { r } 20 x - y = 1 \\x y = 1\end{array} \right.{20x−y=1xy=1​

Solve the Nonlinear System of Equations for Real Solutions $\left\{ \begin{array} { r } 20 x - y = 1 \\x y = 1\end{array} \right.$