If `x -2y = 2 and 3x + y = 20,` then the value of (x,y) is

Options

A. `(6,2)`
B. `(4,1)`
C. `(3,2)`
D. `(5,5)`

Answer

To solve the system of equations given by:

1. \( x - 2y = 2 \) (Equation 1)
2. \( 3x + y = 20 \) (Equation 2)

we can use the elimination method. Here’s a step-by-step solution:

### Step 1: Rewrite the equations
We have the two equations:
- Equation 1: \( x - 2y = 2 \)
- Equation 2: \( 3x + y = 20 \)

### Step 2: Multiply Equation 2
To eliminate \( y \), we can manipulate Equation 2. We will multiply Equation 2 by 2 so that the coefficients of \( y \) in both equations will match:

\[ 2(3x + y) = 2(20) \]

This gives us:

\[ 6x + 2y = 40 \quad \text{(Equation 3)} \]

### Step 3: Add Equation 1 and Equation 3
Now we will add Equation 1 and Equation 3:

\[ (x - 2y) + (6x + 2y) = 2 + 40 \]

This simplifies to:

\[ 7x + 0y = 42 \]

So we have:

\[ 7x = 42 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):

\[ x = \frac{42}{7} = 6 \]

### Step 5: Substitute \( x \) back into Equation 1
Now that we have the value of \( x \), we can substitute it back into Equation 1 to find \( y \):

\[ 6 - 2y = 2 \]

### Step 6: Solve for \( y \)
Rearranging the equation gives us:

\[ -2y = 2 - 6 \]
\[ -2y = -4 \]
\[ y = \frac{-4}{-2} = 2 \]

### Final Result
Thus, the solution to the system of equations is:

\[ (x, y) = (6, 2) \]