Problem 62 - Beginning and Intermediate Algebra

Question

Graph each compound inequality. \(y-2 x \leq 1\) and \(y \geq-\frac{1}{5} x-2\)

Step-by-Step Solution

Answer

Rewrite the inequalities as equalities and plot the lines: \(y - 2x = 1\) is \(y = 2x + 1\), and \(y = -\frac{1}{5}x - 2\). Next, shade the region below the line \(y = 2x + 1\) and the region above the line \(y = -\frac{1}{5}x - 2\). The overlapping shaded region represents the solution to the compound inequality.

Step 1: Rewrite the inequalities as equalities

We will first change the inequalities to equalities and find the corresponding lines. For the first inequality \(y-2x \leq 1\), the equation will be \(y-2x = 1\). For the second inequality \(y \geq -\frac{1}{5}x - 2\), the equation will be \(y = -\frac{1}{5}x - 2\). Now we have two equations to plot the lines on the graph: 1. \(y-2x = 1\) 2. \(y = -\frac{1}{5}x - 2\)

Step 2: Reformat the equations (if needed)

We will rewrite the first equation in slope-intercept form, making it easier to graph. 1. \(y-2x = 1\) can be rewritten as \(y = 2x + 1\).

Step 3: Graph the lines

Plot the two lines on the graph: 1. The line of \(y = 2x + 1\) has a slope of \(2\) and a y-intercept of \((0,1)\). 2. The line of \(y = -\frac{1}{5}x - 2\) has a slope of \(-\frac{1}{5}\) and a y-intercept of \((0,-2)\).

Step 4: Shade the regions for each inequality

Given the inequalities, we need to shade the region of interest for each inequality: 1. For \(y-2x \leq 1\), we know \(y \leq 2x + 1\), meaning that we should shade the region below the line \(y = 2x + 1\). 2. For \(y \geq -\frac{1}{5}x - 2\), we should shade the region above the line \(y = -\frac{1}{5}x - 2\).

Step 5: Determine the overlapping region

Now, we have to find the region that satisfies both inequalities simultaneously. Observation of the graph we made in step 4 will show the area where both shaded regions overlap. That overlapping region is the one that satisfies both the given inequalities and represents the solution to the compound inequality.