Question - Solving a System of Linear Equations

Given the system of equations:

\[\begin{align*} 5x + 4y &= 58 \quad (1)\\ 3x + 7y &= 67 \quad (2) \end{align*}\]

To solve it, we can use the method of substitution or elimination. Here, we will use the elimination method for this solution:

Multiply equation (1) by 3 and equation (2) by 5 to make the coefficients of \( x \) identical:

\[\begin{align*} 3*(5x + 4y) &= 3*58 \\ 5*(3x + 7y) &= 5*67 \end{align*}\]

\[\begin{align*} 15x + 12y &= 174 \quad (3)\\ 15x + 35y &= 335 \quad (4) \end{align*}\]

Subtract equation (3) from equation (4) to eliminate \( x \):

\[\begin{align*} (15x + 35y) - (15x + 12y) &= 335 - 174\\ 23y &= 161 \end{align*}\]

Solving for \( y \):

\[y = \frac{161}{23} = 7\]

Now, substitute \( y = 7 \) into equation (1):

\[5x + 4(7) = 58\]

\[5x + 28 = 58\]

\[5x = 58 - 28\]

\[5x = 30\]

Solving for \( x \):

\[x = \frac{30}{5} = 6\]

Thus, the solution to the system of equations is \( x = 6 \) and \( y = 7 \).