Solve the system of equations:
$\begin{array}{c}x=y-5 \\ -3 x+10 y=-20\end{array}$
The problem presents a system of two linear equations with two variables, \(x\) and \(y\):
\[ \begin{cases} x = y - 5 \ -3x + 10y = -20 \end{cases} \]
Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Since the first equation already isolates \(x\), the substitution method is a highly efficient choice for this system.
We substitute the expression for \(x\) given in the first equation, \(x = y - 5\), into the second equation \(-3x + 10y = -20\).
This yields:
\[-3(y - 5) + 10y = -20\]
Distributing the \(-3\) across the terms inside the parentheses, we get:
\[-3y + 15 + 10y = -20\]
Combining the like terms involving \(y\) (i.e., \(-3y + 10y\)):
\[7y + 15 = -20\]
Subtracting \(15\) from both sides of the equation to isolate the term with \(y\):
\[7y = -20 - 15\]
\[7y = -35\]
Dividing by \(7\), we find the value of \(y\):
\[y = \frac{-35}{7} = -5\]
Now that we have the value \(y = -5\), we substitute it back into the first equation, \(x = y - 5\), to find the corresponding value of \(x\):
\[x = (-5) - 5\]
\[x = -10\]
Thus, the solution to the system is the ordered pair \((x, y) = (-10, -5)\). We can verify this by checking the second equation:
\[-3(-10) + 10(-5) = 30 - 50 = -20\]
The result matches the right side of the second equation, confirming our solution is correct.
Systems of linear equations are fundamental in resource allocation. For example, if a store sells two types of items, say "x" and "y", and you know their price difference and a combined total for a specific sales event, you can use a system of equations to determine the exact number of each item sold. Solving such problems helps businesses track inventory and understand consumer demand effectively.