Solve the system by the substitution method.
-$$\begin{array} { l }
x y = 6 \
x ^ { 2 } + y ^ { 2 } = 37
\end{array}$$
A) $\{ ( 1,6 ) , ( - 1 , - 6 ) , ( 6,1 ) , ( - 6 , - 1 ) \}$
B) $\{ ( 1,6 ) , ( - 1 , - 6 ) , ( 1 , - 6 ) , ( - 1,6 ) \}$
C) $\{ ( 1,6 ) , ( 6,1 ) , ( 1 , - 6 ) , ( 6 , - 1 ) \}$
D) $\{ ( - 1 , - 6 ) , ( - 6 , - 1 ) , ( - 1,6 ) , ( - 6,1 ) \}$

Question

Solve the system by the substitution method.
-$$\begin{array} { l } 
x y = 6 \
x ^ { 2 } + y ^ { 2 } = 37
\end{array}$$
A) $\{ ( 1,6 ) , ( - 1 , - 6 ) , ( 6,1 ) , ( - 6 , - 1 ) \}$
B) $\{ ( 1,6 ) , ( - 1 , - 6 ) , ( 1 , - 6 ) , ( - 1,6 ) \}$
C) $\{ ( 1,6 ) , ( 6,1 ) , ( 1 , - 6 ) , ( 6 , - 1 ) \}$
D) $\{ ( - 1 , - 6 ) , ( - 6 , - 1 ) , ( - 1,6 ) , ( - 6,1 ) \}$

Quizplus · Accepted Answer

To solve the system by substitution, we can express $x$ in terms of $y$ from the first equation: $x = \frac{6}{y}$. Substituting this into the second equation gives $\left(\frac{6}{y}\right)^2 + y^2 = 37$. Simplifying, we get $36/y^2 + y^2 = 37$, which can be rewritten as $y^4 - 37y^2 + 36 = 0$. This can be treated as a quadratic equation in terms of $y^2$, leading to the solutions $y^2 = 1$ and $y^2 = 36$, giving $y = \pm 1$ and $y = \pm 6$. Substituting these values back into $x = 6/y$ gives the pairs $(6,1)$, $(-6,-1)$, $(1,6)$, and $(-1,-6)$, which matches choice A.

Solve the System by the Substitution Method xy=6x2+y2=37\begin{array} { l } x y = 6 \\x ^ { 2 } + y ^ { 2 } = 37\end{array}xy=6x2+y2=37​

Solve the System by the Substitution Method $\begin{array} { l } x y = 6 \\x ^ { 2 } + y ^ { 2 } = 37\end{array}$