Given:
[tex]\begin{array}{l}
2 x-3 y+z=0 \\
3 x+2 y=35 \\
4 y-2 z=14
\end{array}[/tex]
Which of the following is a solution to the given system?
A. (2,3,5)
B. (3, 2, 0)
C. (1, 16, 0)
D. (7,7,7)
Answer (2)
Substitute each candidate solution into the system of equations.
Check if all three equations are satisfied.
(2, 3, 5), (3, 2, 0), and (1, 16, 0) do not satisfy all equations.
(7, 7, 7) satisfies all three equations, so the solution is ( 7 , 7 , 7 ) .
Explanation
Understanding the Problem We are given a system of three linear equations with three unknowns: x, y, and z.
The equations are:
2 x − 3 y + z = 0
3 x + 2 y = 35
4 y − 2 z = 14
We are given four possible solutions (x, y, z): (2, 3, 5), (3, 2, 0), (1, 16, 0), and (7, 7, 7).
Solution Plan Our objective is to determine which of the given points (x, y, z) is a solution to the system of equations. To do this, we will substitute the values of x, y, and z from each candidate solution into the three equations. If all three equations are satisfied by the values, then the candidate is a solution. If any of the equations are not satisfied, then the candidate is not a solution.
Testing (2, 3, 5) Let's test the first candidate solution (2, 3, 5):
Equation 1: 2 ( 2 ) − 3 ( 3 ) + 5 = 4 − 9 + 5 = 0 . This equation is satisfied. Equation 2: 3 ( 2 ) + 2 ( 3 ) = 6 + 6 = 12 . This is not equal to 35, so the second equation is not satisfied. Equation 3: 4 ( 3 ) − 2 ( 5 ) = 12 − 10 = 2 . This is not equal to 14, so the third equation is not satisfied.
Therefore, (2, 3, 5) is not a solution to the system of equations.
Testing (3, 2, 0) Now, let's test the second candidate solution (3, 2, 0):
Equation 1: 2 ( 3 ) − 3 ( 2 ) + 0 = 6 − 6 + 0 = 0 . This equation is satisfied. Equation 2: 3 ( 3 ) + 2 ( 2 ) = 9 + 4 = 13 . This is not equal to 35, so the second equation is not satisfied. Equation 3: 4 ( 2 ) − 2 ( 0 ) = 8 − 0 = 8 . This is not equal to 14, so the third equation is not satisfied.
Therefore, (3, 2, 0) is not a solution to the system of equations.
Testing (1, 16, 0) Next, let's test the third candidate solution (1, 16, 0):
Equation 1: 2 ( 1 ) − 3 ( 16 ) + 0 = 2 − 48 = − 46 . This is not equal to 0, so the first equation is not satisfied. Equation 2: 3 ( 1 ) + 2 ( 16 ) = 3 + 32 = 35 . This equation is satisfied. Equation 3: 4 ( 16 ) − 2 ( 0 ) = 64 − 0 = 64 . This is not equal to 14, so the third equation is not satisfied.
Therefore, (1, 16, 0) is not a solution to the system of equations.
Testing (7, 7, 7) Finally, let's test the fourth candidate solution (7, 7, 7):
Equation 1: 2 ( 7 ) − 3 ( 7 ) + 7 = 14 − 21 + 7 = 0 . This equation is satisfied. Equation 2: 3 ( 7 ) + 2 ( 7 ) = 21 + 14 = 35 . This equation is satisfied. Equation 3: 4 ( 7 ) − 2 ( 7 ) = 28 − 14 = 14 . This equation is satisfied.
Since all three equations are satisfied, (7, 7, 7) is a solution to the system of equations.
Conclusion Therefore, the solution to the given system of equations is (7, 7, 7).
Examples
Systems of equations are used in various fields, such as engineering, economics, and computer science. For example, in electrical engineering, systems of equations can be used to analyze circuits and determine the current and voltage at different points. In economics, they can be used to model supply and demand curves and find equilibrium prices. Understanding how to solve systems of equations is crucial for solving real-world problems in these fields. Imagine you're baking a cake and need to adjust the ingredient quantities based on the number of servings. If you know the ratios of flour, sugar, and eggs for a standard cake, you can set up a system of equations to find the new quantities needed for a larger or smaller cake. This ensures your cake turns out perfectly, no matter the size!
After testing all candidate solutions against the set of equations, the only solution that satisfies all three equations is (7, 7, 7). Therefore, the correct answer is (7, 7, 7).
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