Solve the simultaneous equations:
$\begin{array}{l}
y=3-x \text{ and } x=2
\end{array}$
We are given a system of two linear equations in two variables \(x\) and \(y\):
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. The second equation directly provides the value of \(x\), which simplifies the problem significantly. We can then use substitution to determine the value of \(y\).
Given the second equation \(x = 2\), we substitute this value into the first equation \(y = 3 - x\):
\[y = 3 - (2)\]
Performing the subtraction, we get:
\[y = 1\]
Thus, the pair of values that satisfies both equations is \(x = 2\) and \(y = 1\).
To verify our results, we check if \(x = 2\) and \(y = 1\) satisfy the original equations:
Both equations are satisfied by the solution \(x = 2\) and \(y = 1\). The final values are \(x = \boxed{2}\) and \(y = \boxed{1}\).
Suppose you are budgeting for a small project where the total cost \(y\) must be 3 units, and you have already committed to spending 2 units on a specific item \(x\). The equation \(y = 3 - x\) represents your remaining budget for other items after spending \(x\) on the first. By setting \(x = 2\), you determine that you have exactly \(y = 1\) unit remaining to spend elsewhere. This type of basic system of equations is fundamental in managing finances, scheduling time, or allocating limited resources efficiently.