/*! 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 2 Solve each linear system of equa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 2

Solve each linear system of equations. In addition, for each system, graph the two lines corresponding to the two equations in a single coordinate system and use your graph to explain your solution. $$ \begin{array}{r} 2 x+3 y=6 \\ x-4 y=-4 \end{array} $$

Short Answer

Expert verified
The solution to the system is \( x = \frac{12}{11} \) and \( y = \frac{14}{11} \), where the lines intersect.

Step by step solution

01

Identify the Equations

The given system of equations is:1. \( 2x + 3y = 6 \)2. \( x - 4y = -4 \). These are the two equations that we need to solve simultaneously.
02

Solve Equation 2 for x

Take the second equation: \( x - 4y = -4 \).Rearrange it to express \( x \) in terms of \( y \):\[ x = 4y - 4 \].
03

Substitute Expression for x in Equation 1

Replace \( x \) in the first equation with \( 4y - 4 \):\[ 2(4y - 4) + 3y = 6 \].This leads to the equation \( 8y - 8 + 3y = 6 \).
04

Simplify and Solve for y

Combine like terms:\[ 11y - 8 = 6 \].Add 8 to both sides:\[ 11y = 14 \].Divide both sides by 11:\[ y = \frac{14}{11} \].
05

Substitute y Back into Expression for x

Now use \( y = \frac{14}{11} \) to find \( x \):\[ x = 4 \left( \frac{14}{11} \right) - 4 \].Calculate:\[ x = \frac{56}{11} - \frac{44}{11} = \frac{12}{11} \].
06

State the Solution

The solution to the system is \( x = \frac{12}{11} \) and \( y = \frac{14}{11} \). This means the lines intersect at this point, which is the single solution to the system.
07

Graph the Equations

Graph the equations by converting them to slope-intercept form:1. For equation 1: \( y = -\frac{2}{3}x + 2 \).2. For equation 2: \( y = \frac{1}{4}x + 1 \).Plot both lines on the same coordinate plane. The point where they intersect is \( \left( \frac{12}{11}, \frac{14}{11} \right) \), confirming the solution.

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.

graphing linear equations
When graphing linear equations, you're drawing a line that represents all the solutions to the equation. Each linear equation can be graphed on a coordinate system. Here鈥檚 how you do it:
  • First, find the y-intercept, which is where the line crosses the y-axis. In equation form, it鈥檚 the constant term in the slope-intercept form, or the value of y when x is 0.
  • Next, use the slope, which is the steepness of the line. The slope is the coefficient of x in slope-intercept form, represented as 'm' in the equation \( y = mx + b \).
  • Finally, plot the y-intercept on the graph. Use the slope to find the next point by moving up or down, and left or right. A positive slope means the line rises as it moves right, and a negative slope means it falls.
By connecting these points with a straight line, you graph the equation. Remember, if two lines intersect, the intersection is the solution to the system of equations.
solving simultaneously
Solving systems of equations simultaneously means finding a common solution for both equations where their graphs intersect. There are different ways to solve them, but solving them algebraically is quite efficient.
In the given exercise:
  • You first rearrange one equation to express one variable in terms of the other.
  • Next, substitute this expression into the other equation. This substitution allows you to have one equation with a single variable, making it possible to solve for that variable.
  • Once you find one variable, substitute it back into the rearranged equation to find the second variable.
This method, known as substitution, is powerful because it directly leads you to the point of intersection of the two lines represented by the original equations. If both equations represent the same line, they have infinitely many solutions. If they are parallel, there are no solutions.
slope-intercept form
The slope-intercept form is a way of writing the equation of a line. It is a simplified form that makes it easy to graph the equation and understand the line's properties.
The general form of the slope-intercept equation is \( y = mx + b \), where:
  • 'm' represents the slope. It shows the rate at which y changes for a change in x. It's calculated as \( \frac{\text{rise}}{\text{run}} \).
  • 'b' is the y-intercept. This is where the line crosses the y-axis, or the value of y when x equals 0.
Converting a linear equation to slope-intercept form is straightforward. You rearrange terms to solve for y and simplify the equation. In the step-by-step solution, both equations were converted to this form to facilitate graph plotting and finding the intersection point.
Understanding this form is essential because it helps you quickly graph the line and appreciate its slope and intercept visually.

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

$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] $$ $$ \text { Compute } A B C $$

Let $$ I_{3}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] $$ Show that \(I_{3}=I_{3}^{2}=I_{3}^{3}\).

Find the augmented matrix and use it to solve the system of linear equations. $$ \begin{aligned} 2 x-z &=1 \\ y+3 z &=x-1 \\ x+z &=y-3 \end{aligned} $$

$$ A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & -2 \\ 3 & 2 \end{array}\right] $$ Show that \((A B)^{-1}=B^{-1} A^{-1}\).

$$ A=\left[\begin{array}{rr} -1 & 0 \\ 1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 2 & 3 \\ -1 & 1 \end{array}\right], \quad C=\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] $$ $$ \text { Show that } A C \neq C A \text { . } $$
See all solutions

Recommended explanations on Biology Textbooks

Ecology

Read Explanation

Biological Molecules

Read Explanation

Reproduction

Read Explanation

Communicable Diseases

Read Explanation

Cell Communication

Read Explanation

Biology Experiments

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.