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)
Rewrite the equations in standard form.
Substitute the values from each option into the equations.
Check if the values satisfy all three equations.
The solution that satisfies all three equations 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 also given four possible solutions, and we need to determine which one is correct.
Rewriting the Equations The given equations are:
x + 2 y − 6 = z
3 y − 2 z = 7
4 + 3 x = 2 y − 5 z
We can rewrite these equations in the standard form:
x + 2 y − z = 6
3 y − 2 z = 7
3 x − 2 y + 5 z = − 4
Testing the Options We will test each of the given options by substituting the values of x , y , and z into the three equations to see if they satisfy all of them.
Option A: x = 2 , y = 1 , z = 2
2 + 2 ( 1 ) − 2 = 2 + 2 − 2 = 2 e q 6
3 ( 1 ) − 2 ( 2 ) = 3 − 4 = − 1 e q 7
3 ( 2 ) − 2 ( 1 ) + 5 ( 2 ) = 6 − 2 + 10 = 14 e q − 4
Option B: x = 0 , y = 5 , z = 4
0 + 2 ( 5 ) − 4 = 10 − 4 = 6
3 ( 5 ) − 2 ( 4 ) = 15 − 8 = 7
3 ( 0 ) − 2 ( 5 ) + 5 ( 4 ) = 0 − 10 + 20 = 10 e q − 4
Option C: x = 3 2 , y = 2 3 , z = − 6 23
3 2 + 2 ( 2 3 ) − ( − 6 23 ) = 3 2 + 3 + 6 23 = 6 4 + 18 + 23 = 6 45 = 2 15 = 6
3 ( 2 3 ) − 2 ( − 6 23 ) = 2 9 + 3 23 = 6 27 + 46 = 6 73 = 7
3 ( 3 2 ) − 2 ( 2 3 ) + 5 ( − 6 23 ) = 2 − 3 − 6 115 = − 1 − 6 115 = 6 − 6 − 115 = 6 − 121 = − 4
Option D: x = 4 7 , y = 2 3 , z = − 4 5
4 7 + 2 ( 2 3 ) − ( − 4 5 ) = 4 7 + 3 + 4 5 = 4 12 + 3 = 3 + 3 = 6
3 ( 2 3 ) − 2 ( − 4 5 ) = 2 9 + 4 10 = 4 18 + 4 10 = 4 28 = 7
3 ( 4 7 ) − 2 ( 2 3 ) + 5 ( − 4 5 ) = 4 21 − 3 − 4 25 = 4 21 − 12 − 25 = 4 − 16 = − 4
Finding the Solution After testing each option, we find that only option D satisfies all three equations.
Stating the Answer 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 the motion of objects, and in engineering to design structures. 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 essential in many fields.
The chosen solution that satisfies all three equations is Option D: x = 4 7 , y = 2 3 , z = − 4 5 . By substituting the values into the equations, we found that only this option holds true for all equations. Hence, it is the correct answer.
;
