/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 In Exercises 19 - 44, solve the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 19 - 44, solve the system of linear equations and check any solution algebraically. \( \left\\{\begin{array}{l}x + y + z = 7\\\2x - y + z = 9\\\3x \hspace{1cm} - z = 10\end{array}\right. \)

Short Answer

Expert verified
The solution to the system of equations is \( x = 3.67 \), \( y = 2.33 \), and \( z = 1 \)

Step by step solution

01

Express Variable

Take the third equation and express \( x \) in terms of \( z \) : \( x = \frac{10 + z}{3} \)
02

Substitute Variable

Substitute \( x \) from the third equation into the first and second equations: First equation: \( \frac{10+z}{3} +y +z =7 \)Second equation: \( 2(\frac{10+z}{3}) -y +z =9 \)
03

Simplify Equations

Simplify the first and second equations to solve for \( y \) and \( z \):First equation: \( 10 + 4z + 3y = 21 \) which simplifies to \( 3y = 11 - 4z \) and on further rearrangement \( y = \frac{11 - 4z}{3} \)Second equation: \( \frac{20+ 2z}{3} - y + z = 9 \), substitute \( y = \frac{11 - 4z}{3} \) to solve for z
04

Obtain Value for z

Solve the simplified equation \( \frac{20+2z - 11 + 4z}{3} + z = 9 \), which simplifies to \( z = 1 \)
05

Determine x and y

Substitute \( z = 1 \) into the equations obtained in earlier steps to solve for \( x \) and \( y \): \( x = \frac{10 + 1}{3} = \frac{11}{3} = 3.67 \) and \( y = \frac{11 - 4(1)}{3} = \frac{7}{3} = 2.33
06

Check the Solution

Substitute \( x = 3.67 \), \( y = 2.33 \), and \( z = 1 \) back into the original system of equations to check if they hold true. All equations hold true, so the solution is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Algebraic Methods
Solving systems of linear equations is fundamental in algebra. Algebraic methods include a variety of techniques such as substitution, elimination, and matrix operations, which aim to find the values of variables that satisfy all equations in the system.

Each method has its own process. For example, the substitution method involves rearranging one of the equations to express one variable in terms of the others, and then substituting this expression into the remaining equations. On the other hand, the elimination method manipulates the equations to cancel out one or more variables, allowing for easier solution of the reduced system.

Factors like the complexity of the equations and the number of variables can influence the choice of method. These algebraic methods are powerful tools that enable us to solve not only simple systems but also more complex ones involving more variables or equations.
Substitution Method
The substitution method is particularly useful when one equation in a system can be easily solved for one variable in terms of the others. This method simplifies the process of finding the solution one variable at a time.

Steps to Solve by Substitution

  • Rearrange one equation to isolate one variable.
  • Substitute this variable's expression into the other equations.
  • Solve the new equation for another variable.
  • Continue this process until all variables are found.
  • Back-substitute the found values to solve for the remaining variables.
  • Check the solution by substituting all variables into the original equations.

This method is highly systematic and enables students to approach the problem one step at a time, thus reducing the potential for confusion.
Systems of Equations
A system of equations is a set of two or more equations with the same set of variables. Systems are classified based on the number of solutions they have: no solution, one solution, or infinitely many solutions.

Solving a system of equations means finding the values of the variables that satisfy all equations in the system simultaneously. This can represent a variety of real-world problems, where the solution reflects the point at which various conditions meet.

When improving understanding of systems of equations, it is crucial to grasp when to use which algebraic method and to develop the ability to identify systems that have no solution or infinitely many solutions, as well as to interpret the meaning of the solution in the context of the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 21-32, use a graphing utility to graph the inequality. \( y \le 2^{2x - 0.5} - 7 \)

In Exercises 55-60, use a graphing utility to graph the solution set of the system of inequalities. \( \left\\{\begin{array}{l} y < -x^2 + 2x + 3\\\ y > x^2 - 4x + 3\end{array}\right. \)

In Exercises 13-16, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. Objective function: \( z = 5x + \dfrac{1}{2}y \) Constraints: \( \hspace{1cm} x \ge 0 \) \( \hspace{1cm} y \ge 0 \) \( \dfrac{1}{2}x + y \le 8 \) \( x + \dfrac{1}{2}y \ge 4 \)

In Exercises 7-12, find the minimum and maximum values of the objective function and where they occur, subject to the indicated constraints. (For each exercise, the graph of the region determined by the constraints is provided.) Objective function: \( z = 10x + 7y \) Constraints: \( 0 \le x \le 60 \) \( 0 \le y \le 45 \) \( 5x + 6y \le 420 \)

An animal shelter mixes two brands of dog food. Brand \(\mathrm{X}\) costs \(\$ 25\) per bag and contains two units of nutritional element A, two units of element \(\mathrm{B},\) and two units of element \(\mathrm{C}\) , Brand \(\mathrm{Y}\) costs \(\$ 20\) per bag and contains one unit of nutritional element A, nine units of element \(\mathrm{B},\) and three units of element \(\mathrm{C}\) . The minimum required amounts of nutrients \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{Care}\) 12 units, 36 units, and 24 units, respectively. What is the optimal number of bags of each brand that should be mixed? What is the optimal cost?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.