Solve the simultaneous equations.
$\begin{array}{c}
x+3 y=12 \\
x+y=4
\end{array}$
$x=\quad y=$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
Our objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since both equations involve \(x\) with a coefficient of 1, the elimination method is an efficient approach.
To eliminate \(x\), we subtract the second equation from the first:
\[(x + 3y) - (x + y) = 12 - 4\]
Performing the subtraction:
\[x - x + 3y - y = 8\]
\[2y = 8\]
Dividing by 2, we find:
\[y = 4\]
Now that we have the value of \(y\), we substitute it back into the second equation, \(x + y = 4\):
\[x + 4 = 4\]
Subtracting 4 from both sides, we get:
\[x = 0\]
Finally, we verify our solution \((x, y) = (0, 4)\) by plugging these values into the first equation:
\[0 + 3(4) = 12\]
\[12 = 12\]
The solution is consistent. Therefore, \(x = 0\) and \(y = 4\).
Solving systems of equations is fundamental in various real-life scenarios. For instance, imagine you are managing two different supplies: pens and notebooks. If a pack of 1 notebook and 3 pens costs \(\$12\), and a pack of 1 notebook and 1 pen costs \(\$4\), you can use this algebraic method to determine the individual price of a pen and a notebook. By setting up the equations \(x + 3y = 12\) (where \(x\) is the price of a notebook and \(y\) is the price of a pen) and \(x + y = 4\), you find that notebooks cost \(\$0\) (or are included) and pens cost \(\$4\). This math helps optimize budgets and compare different purchasing options efficiently.