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

From the second equation, $3x - 4y = 0$, we can express $x$ in terms of $y$ or vice versa. Let's express $x$ in terms of $y$: $3x = 4y$, so $x = \frac{4y}{3}$. Substituting $x = \frac{4y}{3}$ into the first equation, $x^2 + y^2 = 25$, we get $\left(\frac{4y}{3}\right)^2 + y^2 = 25$. Solving this equation for $y$, we find $y = 3$ and $y = -3$. Substituting these values back into $x = \frac{4y}{3}$, we find corresponding $x$ values of $4$ and $-4$, respectively. Therefore, the solution set is $(4, 3)$ 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 $\left\{ \begin{array} { l } x ^ { 2 } + y ^ { 2 } = 25 \\3 x - 4 y = 0\end{array} \right.$