Question

1/3x + y = 3 and 2y = -2/3x + 10
solve using the elimination method
show the steps

Answer 1

To solve the system of equations using the elimination method, we first need to rewrite the equations in standard form (Ax + By = C) and then manipulate them to eliminate one variable.

The original equations are:

1. \( \frac{1}{3}x + y = 3 \)
2. \( 2y = -\frac{2}{3}x + 10 \)

### Step 1: Rewrite both equations in standard form

**Equation 1:**
Multiply every term by 3 to eliminate the fraction:

\[
3\left(\frac{1}{3}x\right) + 3y = 3 \times 3
\]
\[
x + 3y = 9 \tag{1}
\]

**Equation 2:**
First, rewrite it in a standard form by moving all terms to one side. We can do this by adding \( \frac{2}{3}x \) to both sides:

\[
2y + \frac{2}{3}x = 10
\]

Now, multiply every term by 3 to eliminate the fraction:

\[
3(2y) + 3\left(\frac{2}{3}x\right) = 3 \times 10
\]
\[
6y + 2x = 30
\]

Rearranging gives:

\[
2x + 6y = 30 \tag{2}
\]

### Step 2: Prepare for elimination

Now we have:

1. \( x + 3y = 9 \)
2. \( 2x + 6y = 30 \)

### Step 3: Use substitution or elimination

We will multiply Equation (1) by 2 to align the coefficients of \(x\) for elimination.

\[
2(x + 3y) = 2 \times 9
\]

This gives us the new equation:

\[
2x + 6y = 18 \tag{3}
\]

### Step 4: Subtract the equations

Now we have:

- Equation (2): \( 2x + 6y = 30 \)
- Equation (3): \( 2x + 6y = 18 \)

Subtract Equation (3) from Equation (2):

\[
(2x + 6y) - (2x + 6y) = 30 - 18
\]
\[
0 = 12
\]

### Conclusion

The result \( 0 = 12 \) is a contradiction. This means that the original system of equations has no solution. Therefore, the two lines represented by the equations are parallel and do not intersect.

### Final Result

The system of equations is inconsistent, indicating that there are no solutions.