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Chapter 9: Problem 5

Graph each inequality. $$x+2 y>4$$

Short Answer

Expert verified
The graph of the inequality \(x+2y > 4\) is a dashed line representing the boundary \(y = -0.5x + 2\), with the region above (greater y-values than) this line shaded to represent the solution to the inequality. A test point confirmed the correct shading.

Step by step solution

01

Convert the inequality into slope-intercept form

The first step will be to manipulate the inequality into slope-intercept form (y=mx+b), that is easier to plot. We'll rewrite \(x+2y > 4\) into this form by first subtracting x from both sides to get: \(2y > -x + 4\). Let鈥檚 isolate y by dividing every term by 2: \(y > -0.5x+2\). Now we have the inequality in slope-intercept form.
02

Graph the boundary line

Now we graph the line \(y = -0.5x + 2\). Since the inequality is 'greater than' and not 'greater than or equal to', we will draw a dashed line to reflect this. If it were 'greater than or equal to', we would have drawn a solid line instead.
03

Choose which region to shade

With the inequality dividing the plane into two regions, we have to determine which region to shade. Since the inequality is \(y > -0.5x + 2\), we shade the region above the line as this represents the values of y that are greater than \(-0.5x + 2\). To confirm this we can use a test point. The origin (0,0) is not included in the inequality, so we can use it as a test point. For the point (0,0), the inequality yields \(0 > 2\) which is not true, so we made the correct decision to shade the region above the line.

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Key Concepts

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

Slope-Intercept Form
When graphing inequalities, the first step is often to convert the inequality into the slope-intercept form, which is written as \(y = mx + b\). This form is very helpful for graphing purposes, as it clearly shows the slope \(m\) and the y-intercept \(b\) of the line.For example, starting with the inequality \(x + 2y > 4\), we rearrange it into slope-intercept form:
  • Subtract \(x\) from both sides to get \(2y > -x + 4\).
  • Isolate \(y\) by dividing each term by 2, resulting in \(y > -0.5x + 2\).
Now we can see that the slope \(m\) of the line is \(-0.5\) and the y-intercept \(b\) is 2. This form makes it straightforward to plot on a graph. The intercept tells you where the line crosses the y-axis, and the slope indicates how steep the line is and its direction.
Boundary Line
After transforming the inequality into slope-intercept form, we need to graph its boundary line. The boundary line quadrants the plane into different regions represented by the inequality.For the inequality \(y > -0.5x + 2\), the boundary line would be the equation \(y = -0.5x + 2\). Depending on the inequality's symbol:
  • A dashed line is used if the inequality is either \(>\) or \(<\), indicating that the points on the line are not included in the region.
  • A solid line is used for \(\geq\) or \(\leq\) to show that points on the line are included in the solution set.
In this specific example, because we have \(y> -0.5x + 2\), we use a dashed line to illustrate the boundary, emphasizing that points on this line are not part of the solution.
Shading Regions
Once the boundary line is plotted, the next step is to determine which region to shade. This highlights which values satisfy the inequality.To figure out which side of the boundary line to shade, we can use a test point that isn't on the line. The origin \((0, 0)\) is a common choice if it doesn't lie on the line:
  • Substitute the test point into the original inequality. For \(y > -0.5x + 2\), substituting \((0, 0)\) gives \(0 > 2\), which is false.
Thus, the region containing the origin is not shaded. Therefore, we fill the opposite side, indicating that the inequality holds in that area. In this case, shade the region above the boundary line because the inequality is \(y > -0.5x + 2\), emphasizing the values of \(y\) that are greater than the values along the boundary line.

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Most popular questions from this chapter

a. Solve: \(x-3<5\) b. Solve: \(2 x+4<14\) c. Give an example of a number that satisfies the inequality in part (a) and the inequality in part (b). d. Give an example of a number that satisfies the inequality in part (a), but not the inequality in part (b).

Without graphing, in Exercises \(55-58,\) determine if each system has no solution or infinitely many solutions. $$\left\\{\begin{array}{l} 3 x+y<9 \\ 3 x+y>9 \end{array}\right.$$

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Rewrite each inequality in the system without absolute value bars. Then graph the rewritten system in rectangular coordinates. \\\ |y| \leq 3\end{array}\right.\end{aligned}$$\left\\{\begin{array}{l}|x| \leq 1 \\ |y| \leq 2\end{array}\right.$
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