Solve by substitution:
$\left\{\begin{array}{l}
4 x+2 y=4 \\
2 x-y=-14
\end{array}\right.$
The objective is to solve the system of two linear equations with two variables \(x\) and \(y\):
\[4x + 2y = 4\]
\[2x - y = -14\]
The substitution method involves isolating one variable from one of the equations and substituting it into the other to reduce the system to a single-variable equation. The second equation, \(2x - y = -14\), is convenient for isolating \(y\).
We begin by isolating \(y\) in the second equation:
\[2x - y = -14\]
Adding \(y\) to both sides and adding \(14\) to both sides, we get:
\[y = 2x + 14\]
Now, we substitute this expression for \(y\) into the first equation:
\[4x + 2(2x + 14) = 4\]
Distribute the \(2\):
\[4x + 4x + 28 = 4\]
Combine like terms:
\[8x + 28 = 4\]
Subtract \(28\) from both sides:
\[8x = 4 - 28\]
\[8x = -24\]
Divide by \(8\):
\[x = -3\]
With the value of \(x = -3\) determined, we substitute it back into our isolated expression for \(y\):
\[y = 2x + 14\]
\[y = 2(-3) + 14\]
\[y = -6 + 14\]
\[y = 8\]
To verify the solution, we check it against the original equations:
Equation 1: \(4(-3) + 2(8) = -12 + 16 = 4\). This holds true.
Equation 2: \(2(-3) - 8 = -6 - 8 = -14\). This also holds true.
Thus, the solution is \(\boxed{x = -3, y = 8}\).
Systems of linear equations like this are frequently used in logistics and economics. For instance, if you are running a business where you have two different shipping options, each with a fixed base fee and a per-package variable cost, you can model the total cost of each option as a linear equation. Solving for the point where the two costs are equal (the intersection) helps you decide which shipping provider is more cost-effective based on the number of packages you are shipping. This algebraic method helps in finding that exact "break-even" point.