Solve the simultaneous equations.
$\begin{aligned}
6 x+y & =15 \\
3 x+y & =3
\end{aligned}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
\[\begin{cases} 6x + y = 15 \ 3x + y = 3 \end{cases}\]
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since both equations have a positive \(y\) term with a coefficient of 1, the elimination method is a highly efficient way to solve this system. By subtracting one equation from the other, we can eliminate the variable \(y\) and solve for \(x\).
To eliminate \(y\), we subtract the second equation from the first:
\[(6x + y) - (3x + y) = 15 - 3\]
Expanding this expression gives:
\[6x - 3x + y - y = 12\]
\[3x = 12\]
Now, divide both sides by 3 to find the value of \(x\):
\[x = \frac{12}{3} = 4\]
Now that we have the value of \(x = 4\), we substitute it into the second equation (\(3x + y = 3\)) to solve for \(y\):
\[3(4) + y = 3\]
\[12 + y = 3\]
Subtract 12 from both sides to isolate \(y\):
\[y = 3 - 12\]
\[y = -9\]
To verify our results, we substitute \(x = 4\) and \(y = -9\) back into both original equations:
Equation 1: \(6(4) + (-9) = 24 - 9 = 15\). This is correct.
Equation 2: \(3(4) + (-9) = 12 - 9 = 3\). This is correct.
Thus, the solution is \(x = 4\) and \(y = -9\). The final answer is \(\boxed{x=4, y=-9}\).
Systems of equations are used frequently in real-world scenarios to determine values that satisfy multiple constraints simultaneously. For example, consider a parking garage that charges a base fee plus an hourly rate. If you know the total cost for two different durations, you can set up a system of equations where \(x\) represents the hourly rate and \(y\) represents the fixed base fee. Solving this system allows you to identify exactly what the garage charges, helping you make informed decisions about where to park.