Solve the simultaneous equations:
$\begin{aligned}
6x + y &= 15 \\
3x + y &= 3
\end{aligned}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
1) \(6x + y = 15\)
2) \(3x + y = 3\)
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since both equations have a term with \(y\) having a coefficient of 1, the elimination method is an efficient approach here.
To eliminate \(y\), we subtract the second equation from the first equation:
\[(6x + y) - (3x + y) = 15 - 3\]
This simplifies to:
\[6x - 3x + y - y = 12\]
\[3x = 12\]
Dividing both sides by 3, we find the value of \(x\):
\[x = \frac{12}{3} = 4\]
Now that we have the value of \(x = 4\), we substitute this back into one of the original equations to solve for \(y\). Using the second equation, \(3x + y = 3\):
\[3(4) + y = 3\]
\[12 + y = 3\]
Subtracting 12 from both sides, we get:
\[y = 3 - 12 = -9\]
To ensure our solution is correct, we verify by substituting \(x = 4\) and \(y = -9\) into both original equations:
1) \(6(4) + (-9) = 24 - 9 = 15\) (This matches the first equation)
2) \(3(4) + (-9) = 12 - 9 = 3\) (This matches the second equation)
Both equations are satisfied, so our solution is correct. Thus, \(x = 4\) and \(y = -9\).
Systems of linear equations are fundamental in many real-world applications, such as optimizing resource allocation or balancing budgets. For example, if you are buying supplies for a project and know the total cost for two different bundles of items, you can use these methods to determine the individual price of each item. By representing the problem as a system of equations, you can quickly find the exact costs or quantities needed to satisfy your constraints, ensuring both efficiency and accuracy in your planning.