Question - Solving a System of Linear Equations Using Elimination Method

To solve the system of equations:

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

We can use the elimination method by multiplying the first equation by 7 and the second equation by -4 to eliminate \(y\):

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

\[\begin{align*} 35x + 28y &= 406 \\ -12x - 28y &= -268 \end{align*}\]

Add the two equations to eliminate \(y\):

\[\begin{align*} (35x + 28y) + (-12x - 28y) &= 406 + (-268) \\ 23x &= 138 \end{align*}\]

Divide by 23 to solve for \(x\):

\[x = \frac{138}{23} = 6\]

Now substitute \(x = 6\) into one of the original equations to find \(y\). We'll use the first equation:

\[5(6) + 4y = 58\]

\[30 + 4y = 58\]

\[4y = 58 - 30\]

\[4y = 28\]

\[y = \frac{28}{4} = 7\]

Therefore, the solution set is \((x, y) = (6, 7)\).