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Chapter 5: Problem 31

Solve the system of equations. If a system does not have one unique solution, determine the number of solutions to the system. $$ \begin{aligned} 3 x+2 y+5 z &=12 \\ 3 y+8 z &=-8 \\ 10 z &=20 \end{aligned} $$

Short Answer

Expert verified
The solution is x = 6, y = -8, and z = 2.

Step by step solution

01

- Solve for z

Start with the simplest equation, which is the third one: \[10z = 20\] Divide both sides by 10 to isolate z: \[ z = \frac{20}{10} = 2\]
02

- Substitute z into the second equation

Now that z is known, substitute z = 2 into the second equation: \[ 3y + 8z = -8 \] Replace z with 2: \[ 3y + 8(2) = -8 \] Simplify: \[ 3y + 16 = -8 \] Subtract 16 from both sides: \[ 3y = -8 - 16 = -24 \] Divide both sides by 3: \[ y = \frac{-24}{3} = -8\]
03

- Substitute y and z into the first equation

Now y is known, substitute y = -8 and z = 2 into the first equation: \[ 3x + 2y + 5z = 12 \] Replace y with -8 and z with 2: \[ 3x + 2(-8) + 5(2) = 12 \] Simplify: \[ 3x - 16 + 10 = 12 \] Combine like terms: \[ 3x - 6 = 12 \] Add 6 to both sides: \[ 3x = 18 \] Divide both sides by 3: \[ x = \frac{18}{3} = 6\]
04

- Verify the solution

To ensure the solution is correct, substitute x = 6, y = -8, and z = 2 back into the original equations: First equation: \[ 3(6) + 2(-8) + 5(2) = 18 - 16 + 10 = 12 \] It holds true. Second equation: \[ 3(-8) + 8(2) = -24 + 16 = -8 \] It holds true. Third equation: \[ 10(2) = 20 \] It holds true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving linear equations
Linear equations are equations where each term is either a constant or the product of a constant and a single variable.
The general form of a linear equation in three variables is:
  • ax + by + cz = d
The first step in solving a system of linear equations is to isolate one of the variables. This often simplifies the expression and helps us find the values of the other variables.
In our exercise, we started by isolating z using the third equation:
  • 10z = 20.
By solving for z, we found that z = 2.
Once one variable is known, it can be substituted back into the other equations to find the remaining unknowns.
substitution method
The substitution method involves solving one equation for one variable, then substituting that expression into another equation.
This method reduces the number of variables in the other equation, making it easier to solve.
For our exercise, after finding z = 2, we substituted this value into the second equation:
  • 3y + 8z = -8.
Replacing z with 2, we simplified to find y = -8.
With both z and y known, we substituted these values into the first equation:
  • 3x + 2y + 5z = 12.
This allowed us to find x = 6.
Substitution is powerful because it progressively reduces the system to simpler forms.
unique solutions
A system of linear equations can have a unique solution, no solution, or infinitely many solutions.
A unique solution means there's exactly one set of values for the variables that satisfies all the equations.
In our exercise, we found unique values for x, y, and z.
  • x = 6
  • y = -8
  • z = 2
These values satisfy all three original equations.
The presence of unique solutions often indicates that the lines or planes represented by the equations intersect at a single point.
verification of solutions
Once we've found values for the variables that we believe solve the system, we need to verify these solutions.
This involves substituting the found values back into the original equations to check that all equations are satisfied simultaneously.
For our exercise, we checked:
  • The first equation:
  • 3(6) + 2(-8) + 5(2) = 18 - 16 + 10 = 12.
  • The second equation:
  • 3(-8) + 8(2) = -24 + 16 = -8.
  • The third equation:
  • 10(2) = 20.
Each statement holds true.
Verification is an essential final step to ensure the correctness of our obtained solutions.

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Most popular questions from this chapter

Josh makes \(\$ 24 /\) hr tutoring chemistry and \(\$ 20 / \mathrm{hr}\) tutoring math. Let \(x\) represent the number of hours per week he spends tutoring chemistry. Let \(y\) represent the number of hours per week he spends tutoring math. a. Write an objective function representing his weekly income for tutoring \(x\) hours of chemistry and \(y\) hours of math. b. The time that Josh devotes to tutoring is limited by the following constraints. Write a system of inequalities representing the constraints. \- The number of hours spent tutoring each subject cannot be negative. \- Due to the academic demands of his own classes he tutors at most \(18 \mathrm{hr}\) per week. \- The tutoring center requires that he tutors math at least 4 hr per week. \- The demand for math tutors is greater than the demand for chemistry tutors. Therefore, the number of hours he spends tutoring math must be at least twice the number of hours he spends tutoring chemistry. c. Graph the system of inequalities represented by the constraints. d. Find the vertices of the feasible region. e. Test the objective function at each vertex. f. How many hours tutoring math and how many hours tutoring chemistry should Josh work to maximize his income? g. What is the maximum income? h. Explain why Josh's maximum income is found at a point on the line \(x+y=18\).

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