/*! 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 12 Solve each system of linear equa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each system of linear equations. $$\begin{array}{r} -y+z=1 \\ x-y+z=-1 \\ x-y-z=-1 \end{array}$$

Short Answer

Expert verified
The solution is \(x = -2\), \(y = -1\), \(z = 0\).

Step by step solution

01

Set up the system of equations

We are given the system of equations: \(-y + z = 1\), \(x - y + z = -1\), and \(x - y - z = -1\). We need to find the values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.
02

Eliminate z from two equations

Add the first and second equations: \(-y + z + x - y + z = 1 - 1\) simplifies to \(x - 2y + 2z = 0\).Next, add the second and third equations:\(x - y + z + x - y - z = -1 - 1\) simplifies to \(2x - 2y = -2\).
03

Simplify the equations

From \(2x - 2y = -2\), divide the entire equation by 2:\(x - y = -1\).Now you have two simplified equations: \(x - 2y + 2z = 0\) and \(x - y = -1\).
04

Solve for x in terms of y

From the simplified equation \(x - y = -1\), solve for \(x\):\(x = y - 1\).
05

Substitute for x in the remaining equation

Substitute \(x = y - 1\) into \(x - 2y + 2z = 0\):\((y - 1) - 2y + 2z = 0\).This simplifies to \(-y + 2z = 1\).
06

Solve for y and z

From \(-y + 2z = 1\), express \(y\) in terms of \(z\):\(y = 2z - 1\).Substitute \(y = 2z - 1\) into the first equation \(-y + z = 1\):\(- (2z - 1) + z = 1\) becomes \(-2z + 1 + z = 1\), simplifying to \(-z + 1 = 1\).Solving for \(z\), we get \(-z = 0\), so \(z = 0\).Substitute \(z = 0\) back to get \(y = 2(0) - 1 = -1\).
07

Solve for x

Substitute the value of \(y = -1\) back into \(x = y - 1\) to get:\(x = -1 - 1 = -2\).
08

Verify the solution

Check the values \(x = -2\), \(y = -1\), \(z = 0\) in the original set of equations.1. \(-y + z = 1\): \(-(-1) + 0 = 1\), which is correct.2. \(x - y + z = -1\): \(-2 - (-1) + 0 = -1\), which is correct.3. \(x - y - z = -1\): \(-2 - (-1) - 0 = -1\), which 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.

Linear Equations
Linear equations are fundamental in algebra and deal with expressions where each term is either a constant or the product of a constant and a single variable. Simply put, linear equations represent straight lines when graphed on a coordinate plane. These equations are called "linear" because they form straight lines. They follow the basic form:
* **Standard form:** \( Ax + By = C \)* **Slope-intercept form:** \( y = mx + b \)* **Point-slope form:** \( y - y_1 = m(x - x_1) \)
In our problem, the system consists of three linear equations, involving three variables: \(x\), \(y\), and \(z\). Our goal is to find a common solution set \((x, y, z)\) that satisfies all these linear equations simultaneously.
The beauty of linear equations is their simplicity and predictability. When solving these systems algebraically, each equation might provide a piece of the puzzle. Together, they lead us to the complete solution.
Solving Equations
To solve equations, especially in a system, mathematical operations are utilized to find the value of unknown variables that make the equation true. In this context, we deal with systems of linear equations, where multiple equations must hold simultaneously.
We often use methods like:
  • Substitution: Solving one equation for a variable and then substituting this expression into another equation.
  • Elimination: Adding or subtracting equations to eliminate one of the variables.
  • Matrix methods: Utilizing row operations to solve equations, especially useful for larger systems.
In our solution: * The elimination method was initially used. By adding different pairs of equations, we eliminated variables systematically. * We transitioned to substitution by solving for one variable in terms of others and subsequently plugging it back. This structured approach eventually leads to the unraveling of each variable, thus solving the system.
Step by Step Solutions
When tackling systems of linear equations, a step by step solution is crucial to ensure clarity and precision. Each phase of the solution contributes crucially to understanding how the final answer is reached.
Here's a brief walkthrough of our exercise:
  • Step 1: Clearly identify your equations. The clarity at this stage determines the efficiency of subsequent steps.
  • Step 2: Look for opportunities to eliminate variables. This simplifies the system into fewer equations with fewer variables.
  • Step 3: Simplify where possible. Reduced and simplified equations make the system more manageable.
  • Step 4-5: Progressively isolate variables, substituting known values to resolve unknowns incrementally.
  • Step 6-7: Continue isolating and substituting until each variable's value is identified.
  • Step 8: Always verify your solution with the original equations to ensure correctness.
Each step carefully builds on the last, revealing the values of \(x\), \(y\), and \(z\) that satisfy the system. This methodical progression is what makes mathematics both logical and satisfying, transforming abstract symbols into concrete solutions.

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 order for \(A_{m \times n}^{2}\) to be defined, what condition (with respect to \(m\) and \(n\) ) must be met?

Use the following tables. The following table gives fuel and electric requirements per mile associated with gasoline and electric automobiles: $$\begin{array}{|l|c|c|}\hline & \begin{array}{c} \text { Numeer of } \\\\\text { GatLons/Mile }\end{array} & \begin{array}{c}\text { Numeer of } \\\\\text { kW-hr/Mile }\end{array} \\\\\hline \text { SUV full size } & 0.06 & 0 \\ \hline \text { Hybrid car } & 0.02 & 0.1 \\\\\hline \text { Electric car } & 0 & 0.3 \\\\\hline\end{array}$$ The following table gives an average cost for gasoline and electricity: $$\begin{array}{|l|c|}\hline \text { Cost per gallon of gasoline } & \$ 3.80 \\\\\hline \text { Cost per kW-hr of electricity } & \$ 0.05 \\\\\hline\end{array}$$ Environment. Let matrix \(A\) represent the gasoline and electricity consumption and matrix \(B\) represent the costs of gasoline and electricity. Find \(A B\) and describe what the elements of the product matrix represent. (Hint: \(A\) has order \(3 \times 2\) and \(B\) has order \(2 \times 1 .)\)

Involve vertical motion and the effect of gravity on an object. Because of gravity, an object that is projected upward will eventually reach a maximum height and then fall to the ground. The equation that relates the height \(h\) of a projectile \(t\) seconds after it is projected upward is given by $$h=\frac{1}{2} a t^{2}+v_{0} t+h_{0}$$ where \(a\) is the acceleration due to gravity, \(h_{0}\) is the initial height of the object at time \(t=0,\) and \(v_{0}\) is the initial velocity of the object at time \(t=0 .\) Note that a projectile follows the path of a parabola opening down, so \(a<0\). An object is thrown upward, and the table below depicts the height of the ball \(t\) seconds after the projectile is released. Find the initial height, initial velocity, and acceleration due to gravity. $$\begin{array}{|c|c|} \hline t \text { (seconos) } & \text { HeiGHT (FEET) } \\ \hline 1 & 34 \\ \hline 2 & 36 \\ \hline 3 & 6 \\ \hline \end{array}$$

In Exercise \(58,\) you were asked to solve this system of equations using an augmented matrix. $$\begin{array}{rr} 2 x+z+y= & -3 \\ 2 y-z+x= & 0 \\ x+y+2 z= & 5 \end{array}$$ A graphing calculator or graphing utility can be use to solve systems of linear equations by entering the coefficients of the matrix. Solve this system and confirm your answer with the calculator's answer.

Evaluate the determinant: $$ \left|\begin{array}{rrrr} 1 & -2 & -1 & 3 \\ 4 & 0 & 1 & 2 \\ 0 & 3 & 2 & 4 \\ 1 & -3 & 5 & -4 \end{array}\right| $$

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.