/*! 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 22 (a) The equation \(x+y=1\) can b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 22

(a) The equation \(x+y=1\) can be viewed as a linear system of one equation in two unknowns. Express a general solution of this equation as a particular solution plus a general solution of the associated homogeneous system. (b) Give a geometric interpretation of the result in part (a).

Short Answer

Expert verified
The general solution is \((x, y) = (1+t, -t)\) and geometrically, it's a line parallel to \(x+y=0\), passing through \((1,0)\).

Step by step solution

01

Understanding the Equation

The given equation is \( x+y = 1 \). This is a linear equation in two variables. We will consider it as a system of equations where only one equation is provided, involving two unknowns \( x \) and \( y \).
02

Solution to the Homogeneous Equation

The associated homogeneous equation is obtained by setting the right-hand side to zero, giving \( x + y = 0 \). This represents a line through the origin where \( y = -x \). The general solution is any point on this line: \((x, -x)\).
03

Find a Particular Solution

Choose specific values for \( x \) and \( y \) that satisfy the original equation \( x + y = 1 \). One simple choice is \( x = 1 \), \( y = 0 \), thus a particular solution is \((1, 0)\).
04

Formulate the General Solution

Combine the particular solution from Step 3 with the general solution of the homogeneous equation from Step 2. Thus, the general solution is \( (x, y) = (1, 0) + (t, -t) \), where \( t \) is a real number denoting parameter.
05

Express the General Solution

Express the derived solutions explicitly: \( x = 1+t \) and \( y = -t \). This represents the general solution for \( t \in \mathbb{R} \).
06

Geometric Interpretation

The original equation \( x+y = 1 \) represents a line in the plane, which is parallel to and one unit above the line \( x+y = 0 \). The general solution found represents every point on this line, shifting along the direction defined by the vector \((1, -1)\). Thus, the line is infinitely extended, defined by the particular point \((1,0)\) and shifted by \((t, -t)\).

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.

General Solution
In the context of linear equations, finding a general solution is a way to express every potential solution to the equation. For the equation \(x + y = 1\), the general solution combines two components: a particular solution and the solution to the associated homogeneous system.
  • A particular solution is a specific set of values for \(x\) and \(y\) that satisfies the equation. For example, choosing \(x = 1\) and \(y = 0\) yields \(x + y = 1\), thus \((1, 0)\) is a particular solution.
  • The general solution of the homogeneous equation derives from \(x + y = 0\), the equation formed by setting the right side of the original equation to zero. This homogeneous equation describes points lying along a line through the origin, specifically \((x, -x)\).
  • By combining these, the general solution for \(x + y = 1\) becomes \((x, y) = (1, 0) + (t, -t)\), where \(t\) can be any real number.
This expression implies that the solution is a line parameterized by \(t\), showing every possible \((x, y)\) pair that satisfies the equation.
Homogeneous System
A homogeneous system in terms of linear equations is a system where all constant terms are zero, such as \(x + y = 0\). Here, each solution of such a system forms a vector space, capturing infinite solutions through vectors.
In the specific case of \(x + y = 0\), this system indicates a line through the origin where \(y\) equals \(-x\), i.e., \((x, -x)\).
  • This line can be described using a parameter \(t\), whereby any point on the line can be denoted as \(\{(t, -t) \mid t \in \mathbb{R}\}\).
  • The solutions form a vector, or a line, through the origin, which plays an essential role in forming a framework for solving non-homogeneous equations.
Understanding homogeneous solutions helps in generalizing solutions for systems of equations, giving a full spectrum of possible solutions beyond any specific values.
Geometric Interpretation
Linear equations can be beautifully visualized in geometric terms. Each equation represents a line in a coordinate space. For \(x + y = 1\), this equation represents a line plotted at exactly one unit above the origin line formed by \(x + y = 0\).
  • The geometric interpretation reveals the nature of the line, how it traverses the plane, specifically parallel to the line represented by \(x + y = 0\).
  • The general solution found \((x, y) = (1, 0) + (t, -t)\) corresponds to a line shifting along a direction vector \((1, -1)\). As \(t\) varies, different points along this line are reached, indicating the line's parallel expansion.
  • The particular solution \((1, 0)\) serves as a specific point on this line, and variations introduced by \(t\) express movement along the plane, respecting the linearity and directionality inherent in the equation.
Through this perspective, linear equations transcend mere algebraic expressions, showcasing their intrinsic connection to geometric spaces.

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 \(\mathbf{u}=(1,-1,3,5)\) and \(\mathbf{v}=(2,1,0,-3) .\) Find scalars \(a\) and \(b\) so that \(a \mathbf{u}+b \mathbf{v}=(1,-4,9,18)\)

(a) What relationship must hold for the point \(\mathbf{p}=(a, b, c)\) to be equidistant from the origin and the \(x z\) -plane? Make sure that the relationship you state is valid for positive and negative values of \(a, b,\) and \(c\). (b) What relationship must hold for the point \(\mathbf{p}=(a, b, c)\) to be farther from the origin than from the \(x z\) -plane? Make sure that the relationship you state is valid for positive and negative values of \(a, b,\) and \(c\).

Find vector and parametric equations of the plane containing the given point and parallel vectors. Point: (-3,1,0)\(;\) vectors: \(\mathbf{v}_{1}=(0,-3,6)\) and \(\mathbf{v}_{2}=(-5,1,2)\)

$$\text { Simplify }(\mathbf{u}+\mathbf{v}) \times(\mathbf{u}-\mathbf{v})$$.

Find the vector component of \(u\) along a and the vector component of \(u\) orthogonal to a. $$\mathbf{u}=(2,0,1), \mathbf{a}=(1,2,3)$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Geometry

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.