Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show that a system consisting of exactly one linear equation can have no solution, one solution, or infinitely many solutions. Give examples.

Short Answer

Expert verified
Single linear equations can have no solution, one solution, or infinitely many solutions.

Step by step solution

01

Understanding Concepts

A single linear equation in two variables, for example, has the form \( ax + by = c \). Depending on its parameters, it can represent different geometric entities such as a line or a point in a coordinate plane.
02

Case Analysis: One Solution

A single equation like \( 2x + 3y = 5 \) generally represents a line. Any specific pair \((x_0, y_0)\) that satisfies the equation is the single intersection point of all possible solutions, thus generating infinite possibilities along that line.
03

Case Analysis: No Solution

An equation like \( 0x + 0y = 5 \) is impossible because adding zero to zero can never equal a non-zero constant. Thus, this represents no points on a plane, hence no solution exists.
04

Case Analysis: Infinitely Many Solutions

In the case of an equation like \( 0x + 0y = 0 \), any \((x, y)\) pair satisfies this equation, hence representing the entire coordinate plane. Thus, there are infinite solutions.
05

Conclusion

When a single linear equation is perfectly satisfied (e.g., \( 0 = 0 \)), it has infinite solutions. If it is never satisfied (e.g., \( 0 = 5 \)), it has no solutions. For other coefficients not leading to trivial contradictions or identities, the equation has exactly one solution as points lying on that line.

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 Vaia!

Key Concepts

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

No Solution System
A system represented by a single linear equation can sometimes have no possible solutions. This occurs when the equation itself is impossible to satisfy. For instance, consider the equation \(0x + 0y = 5\). Here, the terms on the left side add up to zero, and it's set equal to a non-zero number like 5. Since zero can never equal five, no values of \(x\) and \(y\) will ever make this equation true.

In the realm of linear equations, whenever you find that a constant non-zero value is obtained from the sum of zero coefficients, it translates into the equation having no solutions. Such equations don't correspond to any point or line in geometry; they are essentially void of intersection in a coordinate plane.
One Solution System
In some cases, a single linear equation will have exactly one solution, but more accurately, it can represent a scenario where every pair \((x, y)\) following the equation lies on a line. For example, consider the equation \(2x + 3y = 5\). This equation describes a line in a two-dimensional space. Every point on this line is a solution to the equation.

To visualize what having one solution means, imagine finding the specific point \((x_0, y_0)\) that satisfies the equation perfectly when the context is such that only one intersection point is considered due to specific constraints. However, when not restricted to such settings, the line itself represents infinite solution pairs \((x, y)\) as it stretches across the plane, infinitely satisfying the equation along its length.
Infinite Solutions System
An infinite solution system is essentially a linear equation that holds true for countless pairs \((x, y)\). Take the equation \(0x + 0y = 0\), for example. Since it holds true irrespective of the values of \(x\) and \(y\) (as both sides equal zero), every point on the plane is a solution.

When assessing for infinite solutions, look for situations where the equation simplifies to a tautology, such as \(0 = 0\). This signifies that the line of solutions isn't just a line through specific points but covers the whole plane, highlighting the difference between this scenario and equations representing finite lines or points. Such equations can be visualized as having no boundary on a coordinate system, representing "all-encompassing" 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

Write the augmented matrix for each of the following systems of linear equations. a. \(\quad x-3 y=5 \quad\) b. \(x+2 y=0\) \(2 x+y=1\) $$ y=1 $$ c. \(x-y+z=2\) d. \(x+y=1\) $$ \begin{array}{ll} x-z=1 & y+z=0 \\ y+2 x=0 & z-x=2 \end{array} $$

Three Nissans, two Fords, and four Chevrolets can be rented for $$\$ 106$$ per day. At the same rates two Nissans, four Fords, and three Chevrolets cost $$\$ 107$$ per day, whereas four Nissans, three Fords, and two Chevrolets cost $$\$ 102$$ per day. Find the rental rates for all three kinds of cars.

Consider the following statements about a system of linear equations with augmented matrix \(A\). In each case either prove the statement or give an example for which it is false. a. If the system is homogeneous, every solution is trivial. b. If the system has a nontrivial solution, it cannot be homogeneous. c. If there exists a trivial solution, the system is homogeneous. d. If the system is consistent, it must be homogeneous. Now assume that the system is homogeneous. e. If there exists a nontrivial solution, there is no trivial solution. f. If there exists a solution, there are infinitely many solutions. g. If there exist nontrivial solutions, the row-echelon form of \(A\) has a row of zeros. h. If the row-echelon form of \(A\) has a row of zeros, there exist nontrivial solutions. i. If a row operation is applied to the system, the new system is also homogeneous.

Which of the following matrices are in reduced row-echelon form? Which are in row-echelon form? .a. \(\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]\) b. \(\left[\begin{array}{rrrr}2 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0\end{array}\right]\) c. \(\left[\begin{array}{rrrr}1 & -2 & 3 & 5 \\ 0 & 0 & 0 & 1\end{array}\right]\) d. \(\left[\begin{array}{lllll}1 & 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1\end{array}\right]\) e. \(\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) f. \(\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{array}\right]\)

The scores of three players in a tournament have been lost. The only information available is the total of the scores for players 1 and \(2,\) the total for players 2 and \(3,\) and the total for players 3 and 1 . a. Show that the individual scores can be rediscovered. b. Is this possible with four players (knowing the totals for players 1 and 2,2 and 3,3 and \(4,\) and 4 and 1)?

See all solutions

Recommended explanations on Economics 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.

Sign-up for free