Solve the simultaneous equations:
$\begin{array}{l}
y=x+1 \text { and } y=3-x \\
x=\_, y=\_
\end{array}$
We are given a system of two linear equations in two variables, \(x\) and \(y\):
1) \(y = x + 1\)
2) \(y = 3 - x\)
Our objective is to find the point \((x, y)\) that satisfies both equations simultaneously, which geometrically corresponds to the point where these two lines intersect. Since both equations express \(y\) in terms of \(x\), we can use the substitution method.
Since both equations are equal to \(y\), we can set the right-hand sides equal to each other:
\[x + 1 = 3 - x\]
Now, we solve for \(x\). Adding \(x\) to both sides of the equation gives:
\[x + x + 1 = 3\]
\[2x + 1 = 3\]
Subtracting \(1\) from both sides:
\[2x = 2\]
Dividing by \(2\):
\[x = 1\]
Now that we have the value of \(x\), we can substitute it back into one of the original equations to find the corresponding value of \(y\). Using the first equation, \(y = x + 1\):
\[y = 1 + 1 = 2\]
We can verify this with the second equation, \(y = 3 - x\):
\[y = 3 - 1 = 2\]
The values are consistent. Thus, the solution to the system is \(x = 1\) and \(y = 2\).
Systems of equations are used frequently in business and economics to find the 'equilibrium point' where the supply of a product matches the consumer demand. For example, if the demand for a new gadget is described by the equation \(y = 3 - x\) and the supply is described by \(y = x + 1\), the point where they meet—in this case, \(x=1\) and \(y=2\)—represents the price and quantity at which the market is stable. Understanding how to find this intersection helps analysts make informed decisions about pricing and production levels.