Problem 16
Question
Graph each inequality. $$x<3+2 y$$
Step-by-Step Solution
Verified Answer
Graph the line \(y = \frac{1}{2}x - \frac{3}{2}\) as dashed, and shade above it.
1Step 1: Rearrange the Inequality
First, we need to rearrange the inequality to identify the slope and y-intercept for graphing. The initial inequality is \(x < 3 + 2y\). We need to solve for \(y\) to make it easier to graph. Begin by subtracting 3 from both sides to get \(x - 3 < 2y\).
2Step 2: Solve for y
Now divide every term in the inequality \(x - 3 < 2y\) by 2 to solve for \(y\). This gives us \(\frac{x}{2} - \frac{3}{2} < y\). To convert it to the standard form of \(y > \ldots\), we can write it as \(y > \frac{1}{2}x - \frac{3}{2}\).
3Step 3: Graph the Associated Equality
To graph the inequality \(y > \frac{1}{2}x - \frac{3}{2}\), first graph the line \(y = \frac{1}{2}x - \frac{3}{2}\) as a dashed line. The dashed line indicates that points on the line are not included in the solution of the inequality.
4Step 4: Determine the Correct Region
After drawing the dashed line, determine which region \(y > \frac{1}{2}x - \frac{3}{2}\) covers. Choose a test point not on the line, such as (0,0). Substitute into \(y > \frac{1}{2}x - \frac{3}{2}\): \(0 > \frac{1}{2}(0) - \frac{3}{2}\) simplifies to \(0 > -\frac{3}{2}\), which is true. Thus, the region above the line is shaded as part of the solution.
5Step 5: Finalize the Graph
Shade the entire region above the dashed line on the graph. This represents all the solutions \((x, y)\) pairs that satisfy \(y > \frac{1}{2}x - \frac{3}{2}\). Remember, the line itself is not included, as indicated by the line being dashed.
Key Concepts
Rearranging InequalitiesSlope-Intercept FormShading RegionsTesting Points in Inequalities
Rearranging Inequalities
Rearranging inequalities is a crucial step when solving or graphing them. Often, inequalities are not given in a form that is easy to interpret or graph right away. To graph an inequality, it's helpful to rearrange it to solve for one variable in terms of the other. Consider the inequality \(x < 3 + 2y\). To make graphing easier, we need to express it in terms of \(y\). This means isolating \(y\) on one side of the inequality.
- Start by subtracting 3 from both sides: \(x - 3 < 2y\).
- Next, divide every term by 2: \(\frac{x}{2} - \frac{3}{2} < y\).
- Rearrange it to the more common form \(y > \frac{1}{2}x - \frac{3}{2}\), where the terms are clear for graphing.
Slope-Intercept Form
To effectively graph inequalities, you should become familiar with the slope-intercept form of a line, which is typically written as \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
- In our rearranged inequality \(y > \frac{1}{2}x - \frac{3}{2}\), the coefficient \(\frac{1}{2}\) represents the slope \(m\).
- The constant \(-\frac{3}{2}\) is the y-intercept \(b\), indicating where the line crosses the y-axis.
Shading Regions
When graphing inequalities, shading regions is about representing all the solutions that satisfy the inequality. Once you have drawn the line associated with the equation, you need to determine on which side of the line to shade. For \(y > \frac{1}{2}x - \frac{3}{2}\), the line \(y = \frac{1}{2}x - \frac{3}{2}\) is the boundary.
- The dashed line indicates that points on the line are not included in the solution set. This is because the inequality is strictly greater than (\(>\)), not greater than or equal to (\(\geq\)).
- Once the boundary is established, shading the region above the line represents all points \((x, y)\) where the inequality holds true.
Testing Points in Inequalities
Testing points is an effective technique to determine which region in the graph is the solution for an inequality. After plotting the boundary line, you need to decide which side satisfies the inequality. A common strategy is to choose a test point.
- Select a point that is not on the line, such as \((0,0)\), for its simplicity.
- Substitute this point into the inequality. For \(y > \frac{1}{2}x - \frac{3}{2}\), substitute \((0,0)\): \(0 > \frac{1}{2}(0) - \frac{3}{2}\).
- Simplify to see if the inequality holds: \(0 > -\frac{3}{2}\), which is true.
Other exercises in this chapter
Problem 15
Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable \(z\). $$\begin{array}{r} x+3 y+4 z=14 \\ 2 x-3
View solution Problem 15
Solve each system by substitution. $$\begin{aligned}&4 x+5 y=7\\\&9 y=31+2 x\end{aligned}$$
View solution Problem 16
Find the partial fraction decomposition for each rational expression. $$\frac{x+1}{x^{2}(1-x)}$$
View solution Problem 16
For each matrix, find \(A^{-1}\) if it exists. Do not use a calculator. $$A=\left[\begin{array}{ll} 0.8 & -0.3 \\ 0.5 & -0.2 \end{array}\right]$$
View solution