Solve the system of equations:
$\begin{array}{l}
2x + 3y = 32 \\
5x + 4y = 59
\end{array}$
The problem presents a system of two linear equations in two variables:
1) \(2x + 3y = 32\)
2) \(5x + 4y = 59\)
Our objective is to solve this system by eliminating the variable \(x\). The prompt suggests a specific elimination strategy: multiply the second equation by \(2\) and the first equation by \(-5\), then add the results. This approach ensures the coefficient of \(x\) in the resulting equation becomes zero, allowing us to solve for \(y\) directly.
To eliminate \(x\), we first multiply the second equation by \(2\):
\[2 \times (5x + 4y) = 2 \times 59\]
\[10x + 8y = 118\]
Next, we multiply the first equation by \(-5\):
\[-5 \times (2x + 3y) = -5 \times 32\]
\[-10x - 15y = -160\]
Now, we add the two modified equations together:
\[(10x - 10x) + (8y - 15y) = 118 - 160\]
The \(x\) terms cancel out, leaving:
\[-7y = -42\]
Dividing both sides by \(-7\), we find the value of \(y\):
\[y = \frac{-42}{-7} = 6\]
With \(y = 6\) determined, we substitute this value back into the first original equation to solve for \(x\):
\[2x + 3(6) = 32\]
\[2x + 18 = 32\]
Subtracting \(18\) from both sides:
\[2x = 32 - 18\]
\[2x = 14\]
Dividing by \(2\):
\[x = 7\]
Thus, the solution to the system is \(x = 7\) and \(y = 6\).
Systems of linear equations are fundamental in fields like economics, engineering, and logistics. For instance, consider a small business owner purchasing two types of stationery, notebooks and pens. If the owner buys 2 notebooks and 3 pens for \(\$32\), and later buys 5 notebooks and 4 pens for \(\$59\), the cost of each item can be found by setting up a system of equations. By solving for the unit price of each, the business owner can accurately track expenses and manage budgets, ensuring financial balance in their operations.