Solve the simultaneous equations:
$\begin{array}{c}
x=21-3 y \\
4 x+7 y=59
\end{array}$
We are given a system of two linear equations with two variables, \(x\) and \(y\). The equations are:
1) \(x = 21 - 3y\)
2) \(4x + 7y = 59\)
Our objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. The system is well-structured for the substitution method because the first equation already provides an explicit expression for \(x\) in terms of \(y\).
We substitute the expression for \(x\) from the first equation into the second equation:
\[4(21 - 3y) + 7y = 59\]
Now, we expand the expression inside the parentheses:
\[4 \times 21 - 4 \times 3y + 7y = 59\]
\[84 - 12y + 7y = 59\]
We combine the terms involving \(y\):
\[84 - 5y = 59\]
Next, subtract 84 from both sides:
\[-5y = 59 - 84\]
\[-5y = -25\]
Divide both sides by -5:
\[y = \frac{-25}{-5} = 5\]
Now that we have the value \(y = 5\), we substitute it back into the first equation to find \(x\):
\[x = 21 - 3(5)\]
\[x = 21 - 15\]
\[x = 6\]
To verify, check the values in the second equation: \(4(6) + 7(5) = 24 + 35 = 59\). Both equations are satisfied.
Systems of linear equations are widely used in resource allocation and finance. For instance, if you are buying two types of items, say apples and oranges, and you know the total number of items and the total cost, you can represent this as a system of equations. For example, if you bought a total of 10 items for \$20, where apples cost \$3 and oranges cost \$1, you could solve the system \(x + y = 10\) and \(3x + y = 20\) to find out how many of each you purchased. This mathematical approach helps in making informed budgeting decisions.