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Chapter 6: Problem 1

Graph the solution set to the inequality. $$ x \geq y $$

Short Answer

Expert verified
Graph the line \( y = x \) as a solid line and shade the area above it.

Step by step solution

01

Understanding the Inequality

The inequality given is \( x \geq y \). This means that the solution set includes all the points \((x, y)\) where the value of \(x\) is greater than or equal to the value of \(y\). This is a key observation that will help us graph this inequality.
02

Find the Boundary Line

The boundary is when \( x = y \). This forms the line \( y = x \), which is a diagonal line that passes through the origin with a slope of 1. Graph this line using a dashed line if it were a strict inequality. However, since we have \( x \geq y \), the line itself is part of the solution, so we graph it as a solid line.
03

Choose a Test Point

Pick a test point not on the line to determine which side of the line is part of the solution. A common test point is \((0,0)\). Substitute into the inequality: \( 0 \geq 0 \), which is true. This implies that the area containing the origin is a solution area.
04

Shade the Solution Area

Using the understanding that the inequality holds true for \( x \geq y \), shade the region above the line \( y = x \), which includes the line itself. This shaded region represents all the points \((x, y)\) that satisfy the inequality.

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Key Concepts

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

boundary line
When graphing an inequality, identifying the boundary line is a crucial initial step. The boundary line for an inequality like \( x \geq y \) is determined by replacing the inequality sign with an equality, yielding \( x = y \). This equation represents a diagonal line through the origin in the coordinate plane, with a slope of 1.
  • The slope-intercept form of this line is \( y = x \).
  • The intercept, where the line crosses the y-axis, is at the origin, \( (0, 0) \).
This line divides the plane into two regions. Since the inequality \( x \geq y \) includes equality, we draw this line as a solid line. This indicates that any point on the line \( y = x \) itself is also a solution.
Observing and depicting the boundary line correctly sets the stage for accurately identifying and shading the solution region.
test point
A test point is an effective method used to determine which side of the boundary line represents the solutions of the inequality. Once you have drawn the boundary line, select a test point that does not lie on the line. A common choice is the origin, \( (0, 0) \), unless it is part of the boundary line.
  • Substitute the test point coordinates into the original inequality \( x \geq y \).
  • If the inequality holds true with the test point, then the region containing the test point includes solutions to the inequality.
  • If it does not, then the solution lies on the opposite side.
For the inequality \( x \geq y \), substitute \( (0, 0) \): \( 0 \geq 0 \).This is true, indicating that the region containing the origin is part of the solution set. Using a test point is a simple verification process that is invaluable in ensuring accuracy when graphing.
shading solution area
Shading the solution area is the final step in graphing an inequality. For \( x \geq y \), after confirming the correct side with a test point, we shade the region.
The true solution areas are all points \( (x, y) \) satisfying \( x \geq y \). This includes:
  • The entire region above the line \( y = x \), which extends infinitely.
  • The line itself, \( y = x \), since it is part of the solution set due to the inequality symbol "\( \geq \)".
By shading this portion of the graph, you highlight the solution set visually. This shading conveys not only where the inequality holds true but also provides a complete picture of all potential solutions on the plane. Remember, clarity in shading can make a significant difference in understanding and accuracy.

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Most popular questions from this chapter

Find the minimum value of \(C=4 x+2 y\) subject to the following constraints. $$ \begin{aligned} &\begin{array}{r} x+y \geq 3 \\ 2 x+3 y \leq 12 \end{array}\\\ &x \geq 0, y \geq 0 \end{aligned} $$

If possible, find \(A B\) and \(B A\). $$A=\left[\begin{array}{rrr}2 & 1 & -1 \\\0 & 2 & 1 \\\3 & 2 & -1 \end{array}\right], \quad B=\left[\begin{array}{rr}1 & 0 \\\2 & -1 \\\3 & 1\end{array}\right]$$

Use the given \(A\) and \(B\) to evaluate each expression. $$A=\left[\begin{array}{rrr}3 & -2 & 4 \\\5 & 2 & -3 \\\7 & 5 & 4\end{array}\right], B=\left[\begin{array}{rrr}1 & 1 & -5 \\\\-1 & 0 & -7 \\\\-6 & 4 & 3\end{array}\right]$$ $$A B$$

Evaluate the matrix expression. $$3\left[\begin{array}{rrrr}1 & 0 & 3 & -1 \\\0 & 1 & 2 & -1 \\\1 & 0 & -3 & 1\end{array}\right]-4\left[\begin{array}{rrrr}-1 & 0 & 0 & 4 \\\0 & -1 & 3 & 2 \\\2 & 0 & 1 & -1\end{array}\right]$$

Graph the solution set to the system of inequalities. Use the graph to identify one solution. $$ \begin{array}{r} y \geq x^{2} \\ x+y \leq 6 \end{array} $$
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