Solve the system of equations:
$\begin{array}{l}
4x - 3y = 25 \\
5x + 6y = -37
\end{array}$
To solve the linear equation \(2x + 5y = 25\), we observe that this is a linear equation in two variables, which represents a line in the coordinate plane. Since there is only one equation and two variables, there are infinitely many solutions. We can express one variable in terms of the other to describe the set of solutions. Isolating \(y\), we subtract \(2x\) from both sides to get \(5y = 25 - 2x\). Dividing by 5 gives \(y = 5 - \frac{2}{5}x\). Thus, for any real value \(x\), the corresponding \(y\) is \(5 - 0.4x\). The solution is a parameterized set of points \((x, 5 - 0.4x)\).
To solve the system of equations
\[ \begin{cases} 4x - 3y = 25 \quad (1) \ 5x + 6y = -37 \quad (2) \end{cases} \]
we use the method of elimination. To eliminate \(y\), we look for a common coefficient. Notice that the coefficient of \(y\) in the first equation is \(-3\) and in the second it is \(6\). Multiplying the first equation by 2 gives \(8x - 6y = 50 \quad (3)\). Adding equation (3) to equation (2), we get \((8x - 6y) + (5x + 6y) = 50 - 37\), which simplifies to \(13x = 13\), so \(x = 1\). Substituting \(x = 1\) into the first original equation, \(4(1) - 3y = 25\), results in \(4 - 3y = 25\), which means \(-3y = 21\), so \(y = -7\). We verify this solution in the second equation: \(5(1) + 6(-7) = 5 - 42 = -37\), which is correct. Therefore, the solution to the system is \(x=1, y=-7\).
These methods of solving equations and systems are fundamental in fields like engineering and economics. For example, when balancing a household budget, you might have one constraint (like total income) involving multiple expenses (the linear equation), or when comparing two different service plans with fixed fees and variable rates, you solve a system of equations to find the point of equality where both plans cost the same. Understanding these algebraic tools allows you to find precise answers in such real-world decision-making scenarios.