Solve the system by the method of substitution.
$$\left\{ \begin{array} { l }
x ^ { 2 } + y ^ { 2 } = 25 \
3 x - 4 y = 0
\end{array} \right.$$
A) $( 4,3 ) , ( - 4 , - 3 )$
B) $( 4,3 )$
C) $( 4,3 ) , ( - 4,3 ) , ( - 4 , - 3 ) , ( 4 , - 3 )$
D) $( 3,4 ) , ( 3 , - 4 )$
E) no real solution

Question

Solve the system by the method of substitution.
$$\left\{ \begin{array} { l } 
x ^ { 2 } + y ^ { 2 } = 25 \
3 x - 4 y = 0
\end{array} \right.$$
A) $( 4,3 ) , ( - 4 , - 3 )$
B) $( 4,3 )$
C) $( 4,3 ) , ( - 4,3 ) , ( - 4 , - 3 ) , ( 4 , - 3 )$
D) $( 3,4 ) , ( 3 , - 4 )$
E) no real solution

Quizplus · Accepted Answer

Solving the second equation for $x$, we get $x = \frac{4}{3}y$. Substituting this into the first equation gives $\left(\frac{4}{3}y\right)^2 + y^2 = 25$, which simplifies to $y^2 = 9$, giving $y = 3$ or $y = -3$. Substituting these values back into $x = \frac{4}{3}y$ gives $x = 4$ when $y = 3$ and $x = -4$ when $y = -3$, leading to the solutions $(4,3)$ and $(-4,-3)$.

Solve the System by the Method of Substitution {x2+y2=253x−4y=0\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 25 \\3 x - 4 y = 0\end{array} \right.{x2+y2=253x−4y=0​

Solve the System by the Method of Substitution $\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 25 \\3 x - 4 y = 0\end{array} \right.$