Problem 76 - Algebra 1: Concepts and Skills

Question

Graph the system of linear inequalities. $$ \begin{aligned} &x+2 y<3\\\ &x-3 y>1 \end{aligned} $$

Step-by-Step Solution

Answer

First, graph the line \( x + 2y = 3 \) and shade the region below this line. Second, graph the line \(x-3y=1\) and shade the region above this line. Ultimately, highlight the region where the two shaded areas intersect, as this is the solution to the system of inequalities.

Step 1: Graph the first inequality

The first inequality is \(x + 2y < 3\). Start by making it an equation \(x + 2y = 3\) and draw the line. Since the inequality is '<', shade the region below this line because this region contains the points that satisfy this inequality.

Step 2: Graph the second inequality

Now turn to the second inequality \(x - 3y > 1\). Similarly, start with the equation \(x - 3y = 1\) and plot this line. Since the inequality is '>', shade the region above this line as it contains points satisfying this inequality.

Step 3: Find the intersection of the two areas

The solution to the system of inequalities is the region where the shaded areas from the first two steps overlap. Therefore, highlight the intersecting region as it represents all points (x, y) that satisfy both inequalities.

Key Concepts

Systems of Inequalities

When dealing with systems of inequalities, we are looking at more than one inequality at a time and seeking to find where their conditions are simultaneously met. In the case of linear inequalities, we usually have two or more linear inequalities that we plot on the same coordinate system.

To understand systems like these, imagine each inequality as creating a boundary where one side meets its condition, while the other does not. Just like a single linear equation forms a line that divides the plane into two halves, each inequality in a system contributes a division, creating various regions in the coordinate plane. Students are typically asked to find which part of the plane satisfies all inequalities at once - this common area represents the solution set of the system. By analyzing the intersection of all such regions, we can understand the set of all points that solve the entire system.

Inequality Shading

The process of inequality shading is an essential step in visualizing the solution set of an inequality. To 'shade' correctly, one must first convert the inequality into an equation and graph the corresponding line. This line acts as a boundary, and the inequality will tell us which side of the line is included in the solution.

For an inequality with a '<' sign, we shade the area below the line. Conversely, for a '>' sign, we shade the area above the line. Remember, this shading indicates the infinite collection of points that satisfy the inequality. A dash or solid line is also important – a dashed line indicates that points on the line are not included in the solution (this corresponds to '<' or '>'), whereas a solid line implies inclusion (corresponding to '<=' or '>='). Effective shading is crucial for visually grasping which part of the graph represents potential solutions.

Solution Regions

Identifying the solution regions in a graph of two or more inequalities involves locating where the shaded areas from each inequality overlap. This area of overlap shows every point (x, y) that satisfies all the inequalities in the system. The technique of shading mentioned earlier is what makes this visual detection possible.

When looking at the solution region for a system of linear inequalities, you should expect to find a polygonal shape, which can range from a triangle to a more complex polygon. Sometimes, however, the solution area can be unbounded, extending infinitely in one or more directions. In exercises where you graph these inequalities, the final answer is this region of intersection, where students often use a different pattern or color to emphasize it. Always ensure that each step in shading is done carefully to avoid confusion and accurately determine the solution area, especially when dealing with complex systems.