Solve the system by using any method.
-$$\begin{array} { l }
4 = 2 \lambda x \
10 = 2 \lambda y \
x ^ { 2 } + y ^ { 2 } = 29
\end{array}$$
A) $\{ ( 1,1,5 ) , ( - 1 , - 1 , - 5 ) \}$
B) $\{ ( 1,1,2 ) \}$
C) $\{ ( 1,1,2 ) , ( - 1 , - 1 , - 2 ) \}$
D) $\{ ( 2,5,1 ) , ( - 2 , - 5 , - 1 ) \}$

Question

Solve the system by using any method.
-$$\begin{array} { l } 
4 = 2 \lambda x \
10 = 2 \lambda y \
x ^ { 2 } + y ^ { 2 } = 29
\end{array}$$
A) $\{ ( 1,1,5 ) , ( - 1 , - 1 , - 5 ) \}$
B) $\{ ( 1,1,2 ) \}$
C) $\{ ( 1,1,2 ) , ( - 1 , - 1 , - 2 ) \}$
D) $\{ ( 2,5,1 ) , ( - 2 , - 5 , - 1 ) \}$

Quizplus · Accepted Answer

Solving the first two equations for $\lambda$, we get $\lambda = \frac{2}{x}$ and $\lambda = \frac{5}{y}$. Equating these gives $\frac{2}{x} = \frac{5}{y}$, leading to $2y = 5x$. Substituting $y = \frac{5}{2}x$ into the third equation $x^2 + y^2 = 29$, we get $x^2 + \left(\frac{5}{2}x\right)^2 = 29$. Solving this gives $x = \pm 2$, $y = \pm 5$, and $\lambda = \pm 1$. Thus, the solutions are $(2, 5, 1)$ and $(-2, -5, -1)$.

Solve the System by Using Any Method 4=2λx10=2λyx2+y2=29\begin{array} { l } 4 = 2 \lambda x \\10 = 2 \lambda y \\x ^ { 2 } + y ^ { 2 } = 29\end{array}4=2λx10=2λyx2+y2=29​

Solve the System by Using Any Method $\begin{array} { l } 4 = 2 \lambda x \\10 = 2 \lambda y \\x ^ { 2 } + y ^ { 2 } = 29\end{array}$