Solve the linear system:
$\begin{array}{l}
(x+3 y=1) \cdot 5 \\
-5 x+4 y=-24
\end{array}$
The system provided consists of two linear equations in variables \(x\) and \(y\):
\[ \begin{cases} x + 3y = 1 & (1) \ -5x + 4y = -24 & (2) \end{cases} \]
Our objective is to find a unique pair \((x, y)\) that satisfies both equations simultaneously. We can utilize the method of elimination to isolate one variable at a time, which is an efficient approach for linear systems.
To eliminate \(x\), we can align the coefficients of \(x\) in both equations. The first equation has a coefficient of \(1\) for \(x\), and the second equation has a coefficient of \(-5\). By multiplying the first equation by \(5\), the coefficient of \(x\) becomes \(5\), which will cancel out when added to the coefficient \(-5\) in the second equation.
Multiplying equation (1) by \(5\):
\[5(x + 3y) = 5(1) \implies 5x + 15y = 5 \quad (3)\]
Now, we add equation (3) and equation (2) together to eliminate \(x\):
\[(5x + 15y) + (-5x + 4y) = 5 + (-24)\]
\[(5x - 5x) + (15y + 4y) = -19\]
\[19y = -19\]
Dividing both sides by \(19\), we find the value of \(y\):
\[y = -1\]
With the value of \(y = -1\) found, we can substitute it back into the simplest original equation, which is (1), to solve for \(x\):
\[x + 3(-1) = 1\]
\[x - 3 = 1\]
Adding \(3\) to both sides:
\[x = 4\]
Thus, the candidate solution is \((x, y) = (4, -1)\).
To ensure the accuracy of our result, we substitute \(x = 4\) and \(y = -1\) into the second original equation (2):
\[-5(4) + 4(-1) = -20 - 4 = -24\]
The result matches the right side of the second equation, confirming our solution is correct. The final solution is \(x = 4\) and \(y = -1\).
Linear systems like these are essential in fields such as engineering, economics, and data science. For instance, if you were managing a production line, you might have constraints on two different resources (e.g., labor hours and material costs) for producing two different types of goods. Solving a system of equations helps you determine the exact quantities of each good you can produce while perfectly exhausting your available resources. Understanding this method allows for precise planning and resource optimization in real-world scenarios.