Solve the simultaneous equations:
$\begin{array}{c}
x=1-4 y \\
7 x+3 y=57 \\
x=\quad y=
\end{array}$
We are given a system of two linear equations with two variables, \(x\) and \(y\):
1) \(x = 1 - 4y\)
2) \(7x + 3y = 57\)
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since the first equation already expresses \(x\) in terms of \(y\), the substitution method is the most efficient approach. We will substitute the expression for \(x\) from the first equation into the second equation, solve for \(y\), and then use that value to find \(x\).
Substitute \(x = 1 - 4y\) into the second equation:
\[7(1 - 4y) + 3y = 57\]
Distribute the 7 into the parentheses:
\[7 - 28y + 3y = 57\]
Combine the terms containing \(y\):
\[7 - 25y = 57\]
Subtract 7 from both sides:
\[-25y = 57 - 7\]
\[-25y = 50\]
Divide both sides by -25:
\[y = \frac{50}{-25} = -2\]
Now that we have the value of \(y = -2\), we can substitute it back into the first equation to find \(x\):
\[x = 1 - 4(-2)\]
\[x = 1 + 8\]
\[x = 9\]
The solution to the system of equations is \(x = 9\) and \(y = -2\). We can verify this by checking the second equation:
\[7(9) + 3(-2) = 63 - 6 = 57\]
The values satisfy both equations. Thus, \(x = 9\) and \(y = -2\).
Solving systems of equations is crucial in real-world resource allocation. For example, if you are buying two types of supplies, where one item's quantity depends on the other's (like a fixed ratio of components for a project), and you have a total budget or constraint, you can model the scenario as a system of equations to determine exactly how many of each item you can afford or need. For instance, if you have a budget of \(\$57\) for items costing \(\$7\) and \(\$3\) respectively, and a constraint linking their quantities, you can calculate the exact amounts needed to satisfy your requirements perfectly.