Solve the following system of equations:
[tex]\begin{array}{l}
x+2 y-6=z \\
3 y-2 z=7 \\
4+3 x=2 y-5 z
\end{array}[/tex]
A. [tex]x=2, y=1, z=2[/tex]
B. [tex]x=0, y=5, z=4[/tex]
C. [tex]x=\frac{2}{3}, y=\frac{3}{2}, z=-\frac{23}{6}[/tex]
D. [tex]x=\frac{7}{4}, y=\frac{3}{2}, z=-\frac{5}{4}[/tex]
Answer (2)
Substitute the values from option A into the equations and check if they are satisfied.
Substitute the values from option B into the equations and check if they are satisfied.
Substitute the values from option C into the equations and check if they are satisfied.
Substitute the values from option D into the equations and check if they are satisfied. The correct answer is x = 4 7 , y = 2 3 , z = − 4 5 .
Explanation
Understanding the Problem We are given a system of three linear equations with three unknowns: x, y, and z. Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously. We are provided with four possible solutions, and we need to determine which one is correct. The equations are:
x + 2 y − 6 = z
3 y − 2 z = 7
4 + 3 x = 2 y − 5 z
Solution Strategy We will test each of the given options by substituting the values of x, y, and z into the three equations. If all three equations are satisfied, then that option is the solution.
Testing Option A Let's test option A: x = 2 , y = 1 , z = 2
Equation 1: 2 + 2 ( 1 ) − 6 = 2 + 2 − 6 = − 2 = 2 . So, equation 1 is not satisfied. Equation 2: 3 ( 1 ) − 2 ( 2 ) = 3 − 4 = − 1 = 7 . So, equation 2 is not satisfied. Equation 3: 4 + 3 ( 2 ) = 4 + 6 = 10 , and 2 ( 1 ) − 5 ( 2 ) = 2 − 10 = − 8 . So, 10 = − 8 , and equation 3 is not satisfied.
Therefore, option A is incorrect.
Testing Option B Let's test option B: x = 0 , y = 5 , z = 4
Equation 1: 0 + 2 ( 5 ) − 6 = 10 − 6 = 4 = z . So, equation 1 is satisfied. Equation 2: 3 ( 5 ) − 2 ( 4 ) = 15 − 8 = 7 . So, equation 2 is satisfied. Equation 3: 4 + 3 ( 0 ) = 4 , and 2 ( 5 ) − 5 ( 4 ) = 10 − 20 = − 10 . So, 4 = − 10 , and equation 3 is not satisfied.
Therefore, option B is incorrect.
Testing Option C Let's test option C: x = 3 2 , y = 2 3 , z = − 6 23
Equation 1: 3 2 + 2 ( 2 3 ) − 6 = 3 2 + 3 − 6 = 3 2 − 3 = 3 2 − 9 = − 3 7 . Since z = − 6 23 , and − 3 7 = − 6 14 = − 6 23 , equation 1 is not satisfied.
Therefore, option C is incorrect.
Testing Option D Let's test option D: x = 4 7 , y = 2 3 , z = − 4 5
Equation 1: 4 7 + 2 ( 2 3 ) − 6 = 4 7 + 3 − 6 = 4 7 − 3 = 4 7 − 12 = − 4 5 = z . So, equation 1 is satisfied. Equation 2: 3 ( 2 3 ) − 2 ( − 4 5 ) = 2 9 + 4 10 = 2 9 + 2 5 = 2 14 = 7 . So, equation 2 is satisfied. Equation 3: 4 + 3 ( 4 7 ) = 4 + 4 21 = 4 16 + 21 = 4 37 , and 2 ( 2 3 ) − 5 ( − 4 5 ) = 3 + 4 25 = 4 12 + 25 = 4 37 . So, equation 3 is satisfied.
Therefore, option D is correct.
Final Answer After testing all the options, we found that option D satisfies all three equations. Therefore, the solution to the system of equations is x = 4 7 , y = 2 3 , z = − 4 5 .
Examples
Systems of equations are used in various real-world applications, such as in economics to model supply and demand curves, in physics to analyze forces and motion, and in engineering to design structures and circuits. For example, when designing a bridge, engineers need to solve systems of equations to ensure that the bridge can withstand the forces acting on it. Similarly, economists use systems of equations to predict how changes in interest rates will affect the economy. Understanding how to solve systems of equations is therefore a valuable skill in many fields.
After testing all provided options, Option D: x = 4 7 , y = 2 3 , z = − 4 5 was found to satisfy all equations in the system. Options A, B, and C do not satisfy the equations. Therefore, Option D is the correct solution.
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