Problem 19 - Intermediate Algebra

Question

For Problems \(17-32\), indicate the solution set for each system of inequalities by shading the appropriate region. $$ \left(\begin{array}{l} x-3 y<6 \\ x+2 y \geq 4 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution set is the region where the shaded areas above both lines intersect.

Step 1: Rewrite each inequality

Start by rewriting each inequality in slope-intercept form (y = mx + by=mx+b). This will allow us to graph them easily.For the inequality \(x - 3y < 6\):Move \(x\) to the right: \(-3y < -x + 6\)Divide by \(-3\) (and flip the inequality sign): \(y > \frac{1}{3}x - 2\)For \(x + 2y \geq 4\):Move \(x\) to the right: \(2y \geq -x + 4\)Divide by \(2\): \(y \geq -\frac{1}{2}x + 2\)

Step 2: Graph the first inequality

Graph the line \(y = \frac{1}{3}x - 2\) by finding the intercepts or by using the slope and y-intercept. Starting at its intercept \((0, -2)\), use the slope \(\frac{1}{3}\) to find another point. Since the inequality is \(y > \frac{1}{3}x - 2\), shade the region above the line.

Step 3: Graph the second inequality

Graph the line \(y = -\frac{1}{2}x + 2\). Start at its y-intercept \((0, 2)\), then use the slope \(-\frac{1}{2}\) to determine another point. Since the inequality is \(y \geq -\frac{1}{2}x + 2\), the line itself is included, and we should use a solid line. Shade the region above (and including) the line.

Step 4: Identify the solution region

The solution to the system of inequalities is the region where the shaded areas from both inequalities overlap. Examine the graph to see where the shaded areas from Steps 2 and 3 intersect. This region represents the set of solutions for the system of inequalities.

Key Concepts

Graphing Inequalities

To solve a system of inequalities, the first step is to rewrite each inequality in slope-intercept form, which is given by \(y = mx + b\). This format makes it easier to graph because it clearly shows the slope \(m\) and the y-intercept \(b\).
Once re-written, each inequality will correspond to a line on a graph. The line itself might be dashed or solid, depending on the inequality symbol:

• A dashed line is used for inequalities with '<' or '>', indicating that the line itself is not part of the solution set.
• A solid line is drawn for '\(\leq\)' or '\(\geq\)' since points on the line are included in the solution set.

By plotting these lines on a graph, we can visualize the solutions for each inequality.

Solution Set

The solution set of a system of inequalities is the region where the solutions to individual inequalities overlap.
It represents all the possible combinations of \(x\) and \(y\) that satisfy all inequalities in the system simultaneously.
Once the individual inequality graphs are plotted and shaded, you'll look for where these shaded regions intersect.
This common region is the solution set.
Ensuring accuracy when shading is important for locating this area correctly.

Slope-Intercept Form

The slope-intercept form \(y = mx + b\) is crucial because it is simple to understand and easy to use for graphing.
In this form:

• The variable \(m\) represents the slope of the line, which shows how steep the line is and the incremental rate of change between the variables \(x\) and \(y\).
• The variable \(b\) is the y-intercept, the point where the line crosses the y-axis, which provides an anchor for plotting the line.

This form is applied to both inequalities by isolating \(y\) on one side, hence streamlining the graphing process.

Shading Regions

Shading is an essential step in graphing inequalities, as it indicates which side of the inequality line contains solutions.
After converting inequalities to slope-intercept form and drawing the lines, use shading to show where each inequality is true.

• For \(y > mx + b\) (greater than), shade above the line.
• For \(y < mx + b\) (less than), shade below the line.

When multiple inequalities are involved, like in a system, find where all shadings intersect to identify the "feasible" region that satisfies all conditions. This intersection is crucial, as it is the region of valid solutions.