Solve the system of equations:
$\begin{array}{l}3 x+4 y=-3 \\ x+2 y=-1\end{array}$
The problem is to solve a system of two linear equations in two variables:
\[ \begin{cases} 3x + 4y = -3 \ x + 2y = -1 \end{cases} \]
A system of linear equations involves finding the set of values for \(x\) and \(y\) that satisfy both equations simultaneously. Geometrically, this represents the intersection point of two lines on a 2D plane. We have a variety of methods available to solve this, such as substitution, elimination, or matrix methods. Here, we will use the elimination method, which is efficient for this system.
To solve the system using the elimination method, we aim to eliminate one variable by combining the equations.
Observe the coefficients of \(x\) in the equations: they are \(3\) and \(1\). If we multiply the second equation, \(x + 2y = -1\), by \(-3\), the coefficient of \(x\) will become \(-3\), which is the additive inverse of the coefficient in the first equation.
Multiplying the second equation by \(-3\):
\[-3(x + 2y) = -3(-1)\]
\[-3x - 6y = 3\]
Now, add this new equation to the first equation (\(3x + 4y = -3\)):
\[(3x - 3x) + (4y - 6y) = -3 + 3\]
\[0x - 2y = 0\]
\[-2y = 0\]
\[y = 0\]
We have found that \(y = 0\).
Now that we have the value for \(y\), we substitute it back into one of the original equations to find \(x\). Using the second equation, \(x + 2y = -1\):
\[x + 2(0) = -1\]
\[x + 0 = -1\]
\[x = -1\]
The solution to the system is \(x = -1\) and \(y = 0\). We can verify this by checking the first equation:
\[3(-1) + 4(0) = -3 + 0 = -3\]
The solution satisfies both equations.
Linear systems are widely used in economics and business to find equilibrium points. For instance, imagine you are managing a small shop selling two types of items. If you know the total cost of two different bundles of these items, you can set up a system of linear equations to determine the unit price of each item. Solving this system allows business owners to analyze their pricing strategy and ensure they cover their costs effectively while remaining competitive.