Question

Using the elimination method solve the systems of equations 1/3 x + y = 3, 2 y= -2/3 x + 10 show all steps

Answer 1

To solve the system of equations using the elimination method, we start with the two equations given:

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

### Step 1: Eliminate fractions

First, let's eliminate the fractions in both equations to make them easier to work with.

**For Equation 1:**
Multiply the entire equation by 3 to eliminate the fraction:

\[
3 \left( \frac{1}{3}x + y \right) = 3 \cdot 3
\]
\[
x + 3y = 9 \quad \text{(Equation 1 revised)}
\]

**For Equation 2:**
Multiply the entire equation by 3 to eliminate the fraction:

\[
3 \cdot 2y = 3 \left( -\frac{2}{3}x + 10 \right)
\]
\[
6y = -2x + 30 \quad \text{(Equation 2 revised)}
\]

Now we have the revised system of equations:

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

### Step 2: Make the coefficients of \(y\) the same

We can multiply Equation 1 by 2, so the coefficient of \(y\) matches that in Equation 2.

Multiply Equation 1 by 2:
\[
2(x + 3y) = 2 \cdot 9
\]
\[
2x + 6y = 18 \quad \text{(Equation 1 multiplied)}
\]

### Step 3: Set up the new system

Now we have:

1. \( 2x + 6y = 18 \) (New Equation 1 from multiplication)
2. \( 2x + 6y = 30 \) (Original Equation 2 revised)

### Step 4: Subtract the equations

Now we will subtract Equation 1 (New) from Equation 2:

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

Since this statement \(0 = 12\) is a contradiction, it indicates that the system of equations is inconsistent.

### Conclusion

The two equations represent parallel lines and thus do not intersect. Therefore, there are no solutions to this system of equations.