Solve the simultaneous equations.
$\begin{aligned}
x+4 y & =19 \\
x+3 y & =15
\end{aligned}$
The problem provides a system of two linear equations with two variables, \(x\) and \(y\):
1) \(x + 4y = 19\)
2) \(x + 3y = 15\)
To find the solution, we aim to isolate one variable or use a method such as elimination to simplify the system into a single-variable equation. The coefficients of \(x\) in both equations are \(1\), which makes the elimination method highly efficient.
We can eliminate \(x\) by subtracting the second equation from the first equation:
\((x + 4y) - (x + 3y) = 19 - 15\)
Simplifying the left side:
\(x - x + 4y - 3y = 19 - 15\)
\(y = 4\)
Thus, we have found the value of \(y\).
Now that we have the value \(y = 4\), we substitute it back into either of the original equations to solve for \(x\). Choosing the second equation:
\(x + 3(4) = 15\)
\(x + 12 = 15\)
Subtracting \(12\) from both sides, we find:
\(x = 15 - 12 = 3\)
So, the solution for the system is \(x = 3\) and \(y = 4\). We verify this by checking the first equation: \(3 + 4(4) = 3 + 16 = 19\), which is correct.
Solving systems of equations is fundamental in various fields, such as economics or engineering. For example, if you are buying supplies for an event, you might know the total cost for two different bundles of items. Let \(x\) be the fixed delivery fee and \(y\) be the cost per item. If one order of \(x + 4y\) costs \(\$19\) and another order of \(x + 3y\) costs \(\$15\), you can use this method to determine both the delivery fee (\(x = 3\)) and the individual item cost (\(y = 4\)).