To solve the given system of equations using the elimination method, we start with the two equations:
- \(\frac{1}{3}x + y = 3\)
- \(2y = -\frac{2}{3}x + 10\)
Step 1: Rearrange equations for elimination
Let's express both equations in standard form (Ax + By = C).
For the first equation:
Multiply through by 3 to eliminate the fraction:
\[
x + 3y = 9
\]
This can be labeled as Equation (1).
For the second equation:
First, rewrite the second equation:
\[
2y = -\frac{2}{3}x + 10
\]
Multiply through by 3 to eliminate the fraction:
\[
6y = -2x + 30
\]
Now rearranging gives:
\[
2x + 6y = 30
\]
This can be labeled as Equation (2).
Step 2: Set the equations for elimination
We have the system now:
- \(x + 3y = 9\) (Equation 1)
- \(2x + 6y = 30\) (Equation 2)
Notice that Equation (2) is just \(2\) times Equation (1). Thus, we can see they represent the same line, which means every solution of one is also a solution of the other.
Step 3: Solve for one variable
Let's solve Equation (1) for \(y\):
\[
y = 3 - \frac{1}{3}x
\]
This shows that the system has infinitely many solutions along the line described by the original equations.
Step 4: Interpret the result
The system has infinitely many solutions, which can be expressed in parametric form. Setting \(x = t\) (where \(t\) is any real number) gives:
\[
y = 3 - \frac{1}{3}t
\]
Thus, the solution set can be described as:
\[
(x, y) = (t, 3 - \frac{1}{3}t) \quad \text{for any real number } t.
\]
Summary
The system of equations is dependent and has an infinite set of solutions that can be expressed as the line:
\[
y = 3 - \frac{1}{3}x.
\]