Solve Nonlinear Systems By Substitution
-$$\begin{array} { l }
x y = 1 \
48 x - y = 2
\end{array}$$
A) $\left\{ \left( - \frac { 1 } { 8 } , - 8 \right) , \left( \frac { 1 } { 6 } , 6 \right) \right\}$
B) $\left\{ \left( - 8 , - \frac { 1 } { 8 } \right) , \left( 6 , \frac { 1 } { 6 } \right) \right\}$
C) $\{ ( 8 , - 8 ) , ( - 6,6 ) \}$
D) $\left\{ \left( \frac { 1 } { 6 } , 6 \right] \right\}$

Question

Solve Nonlinear Systems By Substitution
-$$\begin{array} { l } 
x y = 1 \
48 x - y = 2
\end{array}$$
A) $\left\{ \left( - \frac { 1 } { 8 } , - 8 \right) , \left( \frac { 1 } { 6 } , 6 \right) \right\}$
B) $\left\{ \left( - 8 , - \frac { 1 } { 8 } \right) , \left( 6 , \frac { 1 } { 6 } \right) \right\}$
C) $\{ ( 8 , - 8 ) , ( - 6,6 ) \}$
D) $\left\{ \left( \frac { 1 } { 6 } , 6 \right] \right\}$

Quizplus · Accepted Answer

From the first equation, $y = \frac{1}{x}$. Substituting $y$ in the second equation gives $48x - \frac{1}{x} = 2$. Multiplying by $x$ to clear the denominator, we get $48x^2 - 1 = 2x$, or $48x^2 - 2x - 1 = 0$. Solving this quadratic equation yields $x = -\frac{1}{8}$ and $x = \frac{1}{6}$. Substituting these values back into $y = \frac{1}{x}$ gives the corresponding $y$ values, resulting in the solution set $\left\{ \left( - \frac { 1 } { 8 } , - 8 \right) , \left( \frac { 1 } { 6 } , 6 \right) \right\}$.

Solve Nonlinear Systems by Substitution - xy=148x−y=2\begin{array} { l } x y = 1 \\48 x - y = 2\end{array}xy=148x−y=2​

Solve Nonlinear Systems by Substitution
- $\begin{array} { l } x y = 1 \\48 x - y = 2\end{array}$