/*! 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 37 Solve each system. State whether... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each system. State whether it is inconsistent or has infinitely many solutions. If the system has infinitely many solutions, write the solution set with y arbitrary. $$\begin{array}{r} 7 x+2 y=6 \\ 14 x+4 y=12 \end{array}$$

Short Answer

Expert verified
Infinitely many solutions: \( \{(x,y) | x = \frac{6 - 2y}{7}, y \text{ is arbitrary} \} \)

Step by step solution

01

Write the system of equations

Given system of linear equations is \[ \begin{array}{r} 7x + 2y = 6 \ 14x + 4y = 12 \end{array} \]
02

Simplify the second equation

Divide the second equation by 2 to simplify it: \[ \frac{14x + 4y}{2} = \frac{12}{2} \ 7x + 2y = 6 \] This is exactly the same as the first equation.
03

Analyze the system of equations

Since both equations are identical (\(7x + 2y = 6\)), they represent the same line. Thus, the system of equations is not inconsistent, and there are infinitely many solutions.
04

Write the solution set with y as arbitrary

To express the solution set with \(y\) as arbitrary, solve for \(x\): \[ 7x + 2y = 6 \ 7x = 6 - 2y \ x = \frac{6 - 2y}{7} \] The solution set can be written as \( \{(x,y) | x = \frac{6 - 2y}{7}, y \text{ is arbitrary} \} \)

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.

Inconsistent Systems
An inconsistent system of equations occurs when there are no solutions because the lines representing the system are parallel and never intersect. Imagine two lines on a graph that never touch each other. These lines will have the same slope but different y-intercepts.
For example, consider the equations:
$$\begin{array}{r} 3x + 2y = 7 \ 3x + 2y = 5 \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} No matter what values you choose for x and y, you can't satisfy both equations alongside one another. If you were to graph these, they would be two parallel lines.
Always look for differing constants after the same linear combination of variables. If you find these, then the system is inconsistent.
Infinitely Many Solutions
Systems with infinitely many solutions occur when the equations describe the same line. This means every point on one line is also a point on the other.
For instance, consider the system:
$$\begin{array}{r} 7x + 2y = 6 \ 14x + 4y = 12 \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} \ \text{} $$
By simplifying the second equation, you get:
$$7x + 2y = 6$$
These two equations are the same. This means every point that satisfies the first equation also satisfies the second equation.
When solving, we set y as arbitrary and solve for x. So for the given system:
  • Express y as any real number.
  • Then solve for x using y.

In this case:
$$ x = \frac{6 - 2y}{7}$$
Therefore, our solution set includes all the points \( (x, y) \) where x and y satisfy this equation.
Linear Equations
Linear equations in two variables are of the form \( ax + by = c \). These graphs represent straight lines on the coordinate plane.
The system of linear equations involves finding the set of values \( x \) and \( y \) that satisfy all equations in the system simultaneously.
They can have:
  • One solution (lines intersect at one point).
  • Infinitely many solutions (lines are the same).
  • No solutions (lines are parallel).

When solving these systems, it's helpful to:
  • Graph the equations to visualize their relationship.
  • Use substitution or combination methods to find solutions.

For instance, in our exercise:
$$\begin{array}{r} 7x + 2y = 6 \ 14x + 4y = 12 \ \text{} \ \text{} \ \text{} \ \text{} $$
We found they describe the same line, indicating infinitely many solutions. Simplifying, substituting, or graphing can help you identify these relationships clearly.

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

Use the determinant theorems to find the value of each determinant. $$\left|\begin{array}{rrr} 9 & 1 & 7 \\ 12 & 5 & 2 \\ 11 & 4 & 3 \end{array}\right|$$

Solve each linear programming problem. Aid to Disaster Victims Earthquake victims in Haiti need medical supplies and bottled water. Each medical kit measures \(1 \mathrm{ft}^{3}\) and weighs 10 lb. Each container of water is also \(1 \mathrm{ft}^{3}\) but weighs \(20 \mathrm{lb}\). The plane can carry only \(80,000\) Ib with a total volume of \(6000 \mathrm{ft}^{3}\) Each medical kit will aid 4 people, and each container of water will serve 10 people. (a) How many of each should be sent to maximize the number of people helped? (b) If each medical kit could aid 6 people instead of 4, how would the results from part (a) change? PICTURE CANT COPY

Given \(A=\left[\begin{array}{rr}4 & -2 \\ 3 & 1\end{array}\right], B=\left[\begin{array}{rr}5 & 1 \\ 0 & -2 \\ 3 & 7\end{array}\right],\) and \(C=\left[\begin{array}{rrr}-5 & 4 & 1 \\ 0 & 3 & 6\end{array}\right],\) find each product when possible. $$C B$$

For the system below, match each determinant in (a)-(d) with its equivalent from choices \(\mathrm{A}-\mathrm{D}\). $$ \begin{aligned} 4 x+3 y-2 z &=1 \\ 7 x-4 y+3 z &=2 \\ -2 x+y-8 z &=0 \end{aligned} $$ \(\begin{array}{lll}\text { (a) } D & \text { (b) } D_{x}\end{array}\) (c) \(D_{y}\) (d) \(D_{z}\) A. \(\left|\begin{array}{rrr}1 & 3 & -2 \\ 2 & -4 & 3 \\ 0 & 1 & -8\end{array}\right|\) B. \(\left|\begin{array}{rrr}4 & 3 & 1 \\ 7 & -4 & 2 \\\ -2 & 1 & 0\end{array}\right|\)C. \(\left|\begin{array}{rrr}4 & 1 & -2 \\ 7 & 2 & 3 \\ -2 & 0 & -8\end{array}\right|\) D. \(\left|\begin{array}{rrr}4 & 3 & -2 \\\ 7 & -4 & 3 \\ -2 & 1 & -8\end{array}\right|\)

Use Cramer's rule to solve each system of equations. If \(D=0,\) use another method to determine the solution set. $$\begin{array}{r} x+y=4 \\ 2 x-y=2 \end{array}$$

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.