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 given equations in standard form.
Use the elimination method to eliminate x from two equations, resulting in two equations with y and z .
Solve for y and z using elimination or substitution.
Substitute the values of y and z back into one of the original equations to solve for x . The solution is x = 4 7 , y = 2 3 , z = − 4 5 , so the final answer is x = 4 7 , y = 2 3 , z = − 4 5 .
Explanation
Analyzing the Problem We are given a system of three linear equations with three unknowns: x, y, and z. Our goal is to solve for x, y, and z. The given equations are:
x + 2 y − 6 = z
3 y − 2 z = 7
4 + 3 x = 2 y − 5 z
Rewriting the Equations First, we rewrite the equations in the standard form:
x + 2 y − z = 6
3 y − 2 z = 7
3 x − 2 y + 5 z = − 4
Eliminating x We can use the elimination method to solve this system. Let's eliminate x from equations 1 and 3. Multiply equation 1 by -3:
− 3 ( x + 2 y − z ) = − 3 ( 6 ) − 3 x − 6 y + 3 z = − 18
Now, add this modified equation to equation 3:
( 3 x − 2 y + 5 z ) + ( − 3 x − 6 y + 3 z ) = − 4 + ( − 18 ) − 8 y + 8 z = − 22
Divide by -2 to simplify:
4 y − 4 z = 11
Solving for y Now we have two equations with y and z:
3 y − 2 z = 7
4 y − 4 z = 11
Multiply the first equation by -2:
− 2 ( 3 y − 2 z ) = − 2 ( 7 ) − 6 y + 4 z = − 14
Add this to the second equation:
( 4 y − 4 z ) + ( − 6 y + 4 z ) = 11 + ( − 14 ) − 2 y = − 3 y = 2 3
Solving for z Substitute y = 2 3 into the equation 3 y − 2 z = 7 :
3 ( 2 3 ) − 2 z = 7 2 9 − 2 z = 7 − 2 z = 7 − 2 9 − 2 z = 2 14 − 9 − 2 z = 2 5 z = − 4 5
Solving for x Substitute y = 2 3 and z = − 4 5 into the equation x + 2 y − z = 6 :
x + 2 ( 2 3 ) − ( − 4 5 ) = 6 x + 3 + 4 5 = 6 x = 6 − 3 − 4 5 x = 3 − 4 5 x = 4 12 − 5 x = 4 7
Final Answer Therefore, the solution is x = 4 7 , y = 2 3 , z = − 4 5 . Comparing this with the given options, we see that option D matches our solution.
Examples
Systems of equations are used in various real-world applications, such as determining the optimal mix of products to maximize profit, balancing chemical equations, and modeling electrical circuits. For instance, a company might use a system of equations to determine how many units of each product to produce in order to maximize profit, given constraints on resources like labor and materials. Understanding how to solve these systems is crucial for making informed decisions in many fields.
The solution to the system of equations is found to be x = 4 7 , y = 2 3 , z = − 4 5 , which matches option D. The method involved rearranging equations, eliminating variables through manipulation, and solving for the remaining unknowns. This systematic approach shows how to handle a system of linear equations effectively.
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