Question

solve 1/3x+y=3,2y=−2/3x+10 using elimination

Answer 1

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

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

Let's first eliminate the fractions by multiplying both equations to make them easier to work with.

### Step 1: Multiply the equations to eliminate fractions

**Equation 1:**
\[
\frac{1}{3}x + y = 3
\]
Multiply by 3:
\[
x + 3y = 9 \quad \text{(Equation 1')}
\]

**Equation 2:**
\[
2y = -\frac{2}{3}x + 10
\]
Multiply by 3:
\[
6y = -2x + 30 \quad \text{(Equation 2')}
\]

Now we have a new system of equations:

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

### Step 2: Align equations for elimination

We'll manipulate Equation 1' so we can eliminate \( y \) by aligning it with Equation 2'.

We can rewrite Equation 1' as:
\[
2(x + 3y) = 2(9)
\]
This gives:
\[
2x + 6y = 18 \quad \text{(Equation 3)}
\]

Now we have:

1. \( 2x + 6y = 18 \) (Equation 3)
2. \( 2x + 6y = 30 \) (Equation 2')

### Step 3: Subtract the equations

Now we can subtract Equation 3 from Equation 2':
\[
(2x + 6y) - (2x + 6y) = 30 - 18
\]
This simplifies to:
\[
0 = 12
\]

### Step 4: Conclusion

Since we came to a false statement, this means that the two equations represent parallel lines and there is no solution to this system of equations. Thus, the system of equations is inconsistent.