Problem 286 - Elementary Algebra

Question

In the following exercises, solve each system by graphing. $$ \left\\{\begin{array}{l} y \geq-\frac{2}{3} x+2 \\ y>2 x-3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution region is above both lines, including the solid line but not the dashed line.

Step 1 - Graph the First Inequality

Graph the equation of the first inequality. Start by converting the inequality into the equation form: \(y = -\frac{2}{3}x + 2\). Plot the y-intercept at (0, 2). Then, use the slope -\(\frac{2}{3}\) to plot another point: Move down 2 units and to the right 3 units from the y-intercept. Draw a solid line connecting these points because the inequality includes \(\geq\). Shade the area above this line.

Step 2 - Graph the Second Inequality

Convert the second inequality to the equation form: \(y = 2x - 3\). Plot the y-intercept at (0, -3). Then, use the slope 2 to plot another point: Move up 2 units and to the right 1 unit from the y-intercept. Draw a dashed line connecting these points because the inequality is \(>\). Shade the area above this line.

Step 3 - Find the Intersection

Identify the region where the shaded areas of both inequalities overlap. This region satisfies both inequalities. The solution to the system corresponds to this overlapping shaded area. Since one inequality includes the boundary (solid line) and the other does not (dashed line), the overlapping region will be above the solid line and above the dashed line without including the dashed boundary.

Key Concepts

Graphing Inequalities

Graphing inequalities involves representing linear inequalities on a coordinate plane. To do this, you start by graphing each inequality as if it were an equation. For example, if you have the inequality \(y \geq -\frac{2}{3}x + 2\), you first graph the equation \(y = -\frac{2}{3}x + 2\). This involves plotting the y-intercept and using the slope to find another point on the line. For inequalities, you must determine whether to draw a solid or dashed line. Use a solid line if the inequality is \(\leq\) or \(\geq\), and a dashed line if it's \(>\) or \(<\).

After drawing the line, you need to determine which side of the line to shade. The shading represents all the solutions to the inequality. Test a point not on the line (usually (0, 0) if it's not on the line) to see if it satisfies the inequality. If it does, shade that side; if not, shade the opposite side.

Slope-Intercept Form

The slope-intercept form is a way of writing the equation of a line. It is given by \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. The slope \(m\) tells you how steep the line is and the direction it goes. A positive slope means the line rises as it moves from left to right, while a negative slope means it falls.

The y-intercept \(b\) tells you where the line crosses the y-axis. For instance, in the equation \(y = 2x - 3\), the slope is 2 and the y-intercept is -3. To graph this, start at the point (0, -3) on the y-axis. Then, use the slope to find another point. Since the slope is 2, go up 2 units and right 1 unit from the y-intercept.

This form makes it easy to plot the line on a coordinate grid and then decide whether to draw a dashed or solid line based on the inequality.

Solution Region

The solution region is the area on the graph where the shaded areas of all inequalities overlap. This region represents all the points that satisfy every inequality in the system. To find it, you graph each inequality and shade the corresponding region.

Imagine you have the inequalities \(y \geq -\frac{2}{3}x + 2\) and \(y > 2x - 3\). First, graph the inequalities separately, shading each region. Remember, the solid line means the boundary is included, and the dashed line means it is not.

Next, look for where both shaded regions intersect. This overlapping area is the solution to the system. It includes points that make both inequalities true. However, since one inequality's boundary is not included, the solution does not include that line but does include the area above both lines. Identifying this region is crucial for understanding how different linear inequalities work together.