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Chapter 7: Problem 7

Solve this system of equations: $$ \left\\{\begin{array}{l} y=|x| \\ y=2.85 \end{array}\right. $$

Short Answer

Expert verified
The solutions are \( x = 2.85 \) and \( x = -2.85 \).

Step by step solution

01

Understand the System of Equations

The system consists of two equations: 1. \( y = |x| \), which describes a V-shaped graph centered at the origin.2. \( y = 2.85 \), a horizontal line crossing the y-axis at \( y = 2.85 \). The solution to the system is where these two equations intersect.
02

Equate the Expressions for y

Since both equations equal \( y \), we can set them equal to each other: \[ |x| = 2.85 \]. This means the positive and negative values of \( x \) that satisfy this equation will be the solutions.
03

Solve for x

The absolute value equation \( |x| = 2.85 \) can be split into two separate equations: 1. \( x = 2.85 \)2. \( x = -2.85 \)We need to consider both cases because of the absolute value.
04

Verify the Solutions

For both solutions, verify by substituting back into each original equation:1. For \( x = 2.85 \), \( y = |2.85| = 2.85 \), and the second equation is \( y = 2.85 \), both match.2. For \( x = -2.85 \), \( y = |-2.85| = 2.85 \), and again \( y = 2.85 \), both match.

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Key Concepts

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

Absolute Value
The absolute value of a number represents its distance from zero on a number line, always yielding a non-negative output. The notation for absolute value is given by two vertical bars surrounding the number, such as \(|x|\). To break it down:
  • If the value inside the absolute value is positive, the output is the same as the input. For example, \(|5| = 5\).
  • If the value inside the absolute value is negative, the output negates the input, making it positive. For instance, \(|-3| = 3\).
When solving absolute value equations like \(|x| = 2.85\), it鈥檚 important to consider both the positive and negative scenarios. This approach ensures both potential solutions, \(x = 2.85\) and \(x = -2.85\), are checked, as both yield the solution \(y = 2.85\). This characteristic of absolute value creates a symmetry about the y-axis in its graphical representation, forming a distinctive V-shape at the origin.
Graphical Intersection
Graphical intersection refers to the point or points where two different graphs meet or cross each other on a coordinate plane. In the context of a system of equations, these intersection points represent the solutions to the system.In this exercise, we have:
  • The equation \(y = |x|\), which graphs as a V-shaped curve. This is due to the nature of the absolute value鈥攑roducing a mirror effect along the y-axis.
  • The equation \(y = 2.85\), shown as a straight horizontal line that crosses the y-axis at \(2.85\).
The solution to the system is found where these graphs intersect. Visually, this process involves looking for points where the graphs touch or overlap. For the given system, the intersection points occur at \( (2.85, 2.85) \) and \( (-2.85, 2.85) \). Each point solves both equations simultaneously, confirming that these are the true solutions of the system.
Linear Equation
A linear equation is a mathematical statement of equality that forms a straight line when graphed. Typically, it takes the form \(y = mx + b\), where \(m\) represents the slope and \(b\) the y-intercept.In the current problem, the linear equation is \(y = 2.85\). This particular equation forms a horizontal line parallel to the x-axis because it does not incorporate an \(x\) term. Key properties of horizontal lines include:
  • The slope is zero, indicating no rise or fall as \(x\) increases.
  • It crosses the y-axis at only one point鈥攖he y-intercept, here \(y = 2.85\).
Understanding these properties helps in visualizing how this line interacts with other graphs鈥攕uch as the V-shape from the absolute value equation鈥攖o find the points of intersection effectively.

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Most popular questions from this chapter

Find the solution to each system. a. \(\left\\{\begin{array}{l}y=3 x-5 \\ y=-2.5 x+9\end{array}\right.\) b. \(\left\\{\begin{array}{l}y=2(x-4)+15 \\ y=15(x+5)-12\end{array}\right.\)

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