Solve 41x + 53y = 135 and 53x + 41y = 147.

Options

A. `x = 1` and `y = 2`
B. `x = 2` and `y = 1`
C. `x = 2` and `y = 2`
D. `x = 1` and `y = 1`

Explanation

To solve the system of equations given by:

1. \( 41x + 53y = 135 \) (Equation 1)
2. \( 53x + 41y = 147 \) (Equation 2)

we can use the method of elimination or substitution. Here, we'll use the elimination method.

### Step 1: Multiply the equations to align coefficients
To eliminate one of the variables, we can multiply both equations by suitable numbers so that the coefficients of \( y \) become the same. We can multiply Equation 1 by 41 and Equation 2 by 53.

\[ 41(41x + 53y) = 41(135) \implies 1681x + 2173y = 5535 \quad \text{(Equation 3)} \]
\[ 53(53x + 41y) = 53(147) \implies 2809x + 2173y = 7791 \quad \text{(Equation 4)} \]

### Step 2: Subtract the equations
Now, we can subtract Equation 3 from Equation 4 to eliminate \( y \):

\[ (2809x + 2173y) - (1681x + 2173y) = 7791 - 5535 \]

This simplifies to:

\[ 2809x - 1681x = 7791 - 5535 \]
\[ 1128x = 2256 \]

### Step 3: Solve for \( x \)
Now, divide both sides by 1128 to find \( x \):

\[ x = \frac{2256}{1128} = 2 \]

### Step 4: Substitute \( x \) back into one of the original equations
Now that we have \( x = 2 \), we can substitute this value back into either of the original equations to find \( y \). We'll use Equation 1:

\[ 41(2) + 53y = 135 \]
\[ 82 + 53y = 135 \]
\[ 53y = 135 - 82 \]
\[ 53y = 53 \]
\[ y = \frac{53}{53} = 1 \]

### Step 5: Final solution
Thus, the solution to the system of equations is:

\[ x = 2, \quad y = 1 \]