Solve the simultaneous equations:
$\begin{array}{l}
x+4 y=20 \\
x+y=11
\end{array}$
The problem presents a system of two linear equations with two variables, \(x\) and \(y\):
\[x + 4y = 20\]
\[x + y = 11\]
To find the unique pair \((x, y)\) that satisfies both equations, we observe that the coefficients of \(x\) are identical in both equations, which makes the elimination method highly efficient. We can eliminate \(x\) by subtracting one equation from the other.
We begin by subtracting the second equation from the first:
\[(x + 4y) - (x + y) = 20 - 11\]
Simplifying the left side, the \(x\) terms cancel out, leaving:
\[(4y - y) = 9\]
\[3y = 9\]
Dividing both sides by 3, we find the value for \(y\):
\[y = \frac{9}{3} = 3\]
Now that we have the value \(y = 3\), we substitute this into the second equation (\(x + y = 11\)) to determine \(x\):
\[x + 3 = 11\]
Subtracting 3 from both sides, we get:
\[x = 11 - 3 = 8\]
Thus, the solution to the system is \(x = 8\) and \(y = 3\).
To ensure our solution is correct, we verify these values in the first equation (\(x + 4y = 20\)):
\[8 + 4(3) = 8 + 12 = 20\]
Since the result matches the equation, the solution is verified. The final values are \(x = 8\) and \(y = 3\).
Systems of linear equations are fundamental in balancing various real-world situations, such as managing personal budgets. For instance, if you purchase a set number of items at two different prices and know the total number of items and the total cost, you can use a system of equations to determine how many of each item you bought. If you bought 11 pieces of fruit (apples and oranges) for a total of \$20, where apples cost \$1 and oranges cost \$4 each, you can represent the purchase as the system of equations solved above to find that you bought 8 apples and 3 oranges.