To solve the given system of equations by the substitution method, follow these steps:
1. Given System of Equations:
[tex]\[
\begin{cases}
4x + 3y = 0 \quad \text{(Equation 1)} \\
x - 4y = 0 \quad \text{(Equation 2)}
\end{cases}
\][/tex]
2. Solve one equation for one variable:
Let's solve Equation 2 for [tex]\( x \)[/tex]:
[tex]\[
x - 4y = 0 \implies x = 4y
\][/tex]
3. Substitute the expression for [tex]\( x \)[/tex] into the other equation:
Substitute [tex]\( x = 4y \)[/tex] into Equation 1:
[tex]\[
4(4y) + 3y = 0
\][/tex]
4. Simplify and solve for [tex]\( y \)[/tex]:
[tex]\[
16y + 3y = 0 \implies 19y = 0 \implies y = 0
\][/tex]
5. Substitute [tex]\( y \)[/tex] back into the expression for [tex]\( x \)[/tex]:
Substitute [tex]\( y = 0 \)[/tex] into [tex]\( x = 4y \)[/tex]:
[tex]\[
x = 4(0) = 0
\][/tex]
6. State the solution set:
The solution to the system is the ordered pair [tex]\((x, y) = (0, 0)\)[/tex].
Therefore, the correct choice is:
A. The solution set is [tex]\(\{(0, 0)\}\)[/tex].
