/*! 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 3 Solve each system in Exercises \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 3

Solve each system in Exercises \(1-4\) by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure described in this section. Find the point \(\left(x_{1}, x_{2}\right)\) that lies on the line \(x_{1}+5 x_{2}=7\) and on the line \(x_{1}-2 x_{2}=-2 .\) See the figure.

Short Answer

Expert verified
The point of intersection is \(\left(\frac{4}{7}, \frac{9}{7}\right)\).

Step by step solution

01

Write the System of Equations

The given lines are represented by the following equations:1. \(x_1 + 5x_2 = 7\)2. \(x_1 - 2x_2 = -2\).
02

Form the Augmented Matrix

Translate the system of equations into an augmented matrix. The augmented matrix for the system is:\[\begin{bmatrix}1 & 5 & | & 7 \1 & -2 & | & -2 \\end{bmatrix}\]
03

Perform Row Operations to Simplify

To eliminate \(x_1\) from the second equation, perform the operation: Row 2 = Row 2 - Row 1.\[\begin{bmatrix}1 & 5 & | & 7 \0 & -7 & | & -9 \\end{bmatrix}\]
04

Solve for \(x_2\)

From the second row of the matrix, which corresponds to the equation \(-7x_2 = -9\), solve for \(x_2\):\[x_2 = \frac{9}{7}\]
05

Substitute Back to Find \(x_1\)

Substitute the value of \(x_2\) into the first equation \(x_1 + 5x_2 = 7\):\[x_1 + 5\left(\frac{9}{7}\right) = 7\]\[x_1 + \frac{45}{7} = 7\]\[x_1 = 7 - \frac{45}{7}\]\[x_1 = \frac{49}{7} - \frac{45}{7} = \frac{4}{7}\]
06

Write the Solution

The solution to the system, which is the point of intersection of the lines, is \[\left(x_1, x_2\right) = \left(\frac{4}{7}, \frac{9}{7}\right)\]

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 91影视!

Key Concepts

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

Elementary Row Operations
Elementary row operations are essential techniques in linear algebra used to manipulate matrices in solving systems of linear equations. These operations help simplify matrices, making it easier to find solutions. There are three types of elementary row operations:
  • Swapping two rows
  • Multiplying a row by a non-zero scalar
  • Adding or subtracting a multiple of one row to another row
They do not change the solution set of the matrix, which ensures that if a matrix is equivalent to another through these operations, then both represent the same system of equations.
Augmented Matrix
An augmented matrix is a compact way of writing a system of linear equations. It combines both the coefficients of the variables and the constants from the equations into a single matrix.
In our system, we translated the set of equations:
  • Equation 1: \(x_1 + 5x_2 = 7\)
  • Equation 2: \(x_1 - 2x_2 = -2\)
These are represented as:\[\begin{bmatrix} 1 & 5 & | & 7 \ 1 & -2 & | & -2 \end{bmatrix}\]The vertical line separates the coefficients from the constant terms, indicating that each row represents an equation from the system.
System of Linear Equations
A system of linear equations consists of multiple linear equations that involve the same set of variables. The primary goal is to find values for these variables that satisfy all the given equations simultaneously.
For example, our system has two equations:
  • \(x_1 + 5x_2 = 7\)
  • \(x_1 - 2x_2 = -2\)
The intersection of solutions of these equations, if it exists, can typically be found using various methods such as substitution, elimination, or matrix methods like Gaussian elimination.
Intersection of Lines
The intersection of lines in coordinate geometry occurs where two lines meet or cross each other in a plane. For systems of linear equations, each equation represents a line, and solving the system gives the intersection point if the lines are not parallel.
In our exercise, the solution of the system leads us to the coordinates \( \left(x_1, x_2\right) = \left(\frac{4}{7}, \frac{9}{7}\right) \). This point is where both lines, represented by the given equations, meet in the Cartesian plane.
  • The lines have exactly one point of intersection, showing that the system is consistent and independent.
  • If the lines did not meet, the system would either have no solution (parallel lines) or infinitely many solutions (coincident lines).
Understanding the intersection helps visualize the solution of the system in geometric terms.

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

Let \(A\) be an \(m \times n\) matrix, and let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(\mathbb{R}^{n}\) with the property that \(A \mathbf{u}=\mathbf{0}\) and \(A \mathbf{v}=\mathbf{0} .\) Explain why \(A(\mathbf{u}+\mathbf{v})\) must be the zero vector. Then explain why \(A(c \mathbf{u}+d \mathbf{v})=\mathbf{0}\) for each pair of scalars \(c\) and \(d\)

Each statement in Exercises 33鈥�38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.) If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\) are in \(\mathbb{R}^{4}\) and \(\mathbf{v}_{3}=\mathbf{0},\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}, \mathbf{v}_{4}\right\\}\) is linearly dependent.

Find the value(s) of \(h\) for which the vectors are linearly dependent. Justify each answer. \(\left[\begin{array}{r}{1} \\ {-1} \\\ {4}\end{array}\right],\left[\begin{array}{r}{3} \\ {-5} \\\ {7}\end{array}\right],\left[\begin{array}{r}{-1} \\ {5} \\\ {h}\end{array}\right]\)

Each statement in Exercises 33鈥�38 is either true (in all cases) or false (for at least one example). If false, construct a specific example to show that the statement is not always true. Such an example is called a counterexample to the statement. If a statement is true, give a justification. (One specific example cannot explain why a statement is always true. You will have to do more work here than in Exercises 21 and 22.) If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{4}\) are linearly independent vectors in \(\mathbb{R}^{4},\) then \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\\}\) is also linearly independent. [Hint: Think about \(x_{1} \mathbf{v}_{1}+x_{2} \mathbf{v}_{2}+x_{3} \mathbf{v}_{3}+0 \cdot \mathbf{v}_{4}=\mathbf{0} . ]\)

Construct \(3 \times 2\) matrices \(A\) and \(B\) such that \(A \mathbf{x}=\mathbf{0}\) has only the trivial solution and \(B \mathbf{x}=\mathbf{0}\) has a nontrivial solution.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
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.