Verified Answer :
- We define the unknowns as \(x\), \(y\), and \(z\), leading to a system of equations: \(x + y + z = 10\), \(2y - x = 12\), and \(x - y + 2z = 7\).
- We use substitution to express \(x\) and \(z\) in terms of \(y\), resulting in \(x = 2y - 12\) and \(z = 22 - 3y\).
- Solving for \(y\) in the third equation gives \(y = 5\), which leads to \(x = -2\) and \(z = 7\).
- The correct numbers are \(\boxed{-2, 5, 7}\).
Explanation
- Formulating the System of Equations
We represent the three unknown numbers as \(x, y,\) and \(z\). Based on the information provided, we can establish a system of three linear equations:
- The sum of the three numbers is \(10\):
\[x + y + z = 10\]
- Two times the second number minus the first number is equal to \(12\):
\[2y - x = 12\]
- The first number minus the second number plus twice the third number equals \(7\):
\[x - y + 2z = 7\]
- Solving for Variables using Substitution
We begin by isolating \(x\) in the second equation:
\[-x + 2y = 12 \implies x = 2y - 12\]
Next, we substitute this expression for \(x\) into the first equation:
\[(2y - 12) + y + z = 10\]
\[3y - 12 + z = 10\]
\[z = 22 - 3y\]
Now we substitute both \(x = 2y - 12\) and \(z = 22 - 3y\) into the third equation to solve for \(y\):
\[(2y - 12) - y + 2(22 - 3y) = 7\]
\[y - 12 + 44 - 6y = 7\]
\[-5y + 32 = 7\]
\[-5y = -25\]
\[y = 5\]
- Verifying the Results
Now that we have the value of \(y\), we can find \(x\) and \(z\):
Using \(x = 2y - 12\):
\[x = 2(5) - 12 = 10 - 12 = -2\]
Using \(z = 22 - 3y\):
\[z = 22 - 3(5) = 22 - 15 = 7\]
The solution to the system is \(x = -2, y = 5, z = 7\). Checking our results against the original equations:
- \(-2 + 5 + 7 = 10\) (Correct)
- \(2(5) - (-2) = 10 + 2 = 12\) (Correct)
- \(-2 - 5 + 2(7) = -7 + 14 = 7\) (Correct)
The claim that the numbers were \(-9, 1.5,\) and \(17.5\) was incorrect, as these values do not satisfy the third condition. The correct numbers, in increasing order, are \(-2, 5,\) and \(7\).
Examples
Imagine a scenario where a teacher is assigning grades based on three different components: a test, a project, and participation. If the total points across these components are constrained by specific rules—such as the sum of all points needing to equal a certain value, and the weighted relationships between individual components being fixed—systems of linear equations provide an essential tool for identifying the exact points earned for each category. This ensures grading consistency and fairness across all students, providing a systematic way to verify that all constraints are met.