Problem 29
Question
Graph the solution set of each system of inequalities or indicate that the system has no solution. $$ \left\\{\begin{array}{l} {2 x-5 y \leq 10} \\ {3 x-2 y>6} \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution set is the intersecting area on the graph formed after shading the areas under the line for the first inequality and above the line for the second inequality. If there's no intersecting area, then the system has no solution.
1Step 1: Graph the first inequality
The first inequality can be rewritten from y-term perspective: \(2x - 10 = 5y\), thus, \(y = (2/5)x - 2\). It's a straight line with slope \(2/5\) and y-intercept \(-2\). Because of the less than or equal to do ‘<=‘ sign in the inequality, this line will be solid (including its points), and we'll shade the area below it, since the inequality sign points towards y.
2Step 2: Graph the second inequality
For the second inequality, we can rewrite it as \(3x - 6 = 2y\), so \(y = (3/2)x - 3\). It is also a straight line with slope \(3/2\) and y-intercept \(-3\). As the inequality sign here is '>' (greater than), we will use a dashed line (excluding points on the line), and we will shade the area above this line, as the inequality sign points away from y.
3Step 3: Determining the solution set
The solution for the system of inequalities is the intersection of the shaded areas from both inequalities. If there's no intersection, then the system has no solution. If there's an intersection, then we mark that area as our solution.
Key Concepts
Graphing InequalitiesIntersection of InequalitiesLinear Inequality Solutions
Graphing Inequalities
Graphing inequalities is like graphing lines, except instead of just a line, you also shade an area to represent all possible solutions. When you have an inequality in the form of either "greater than" or "less than," you can represent it visually on a graph to determine which values are valid. Here’s a simple way to approach graphing:
- Rearrange the inequality to the "y = mx + b" form to identify the slope and y-intercept.
- Plot the line as usual. If the inequality is "less than or equal to" or "greater than or equal to," draw a solid line. If it’s "less than" or "greater than," draw a dashed line to show the line itself isn't included in the solution.
- Shade the area above or below the line based on the inequality. For "y >" or "y ≥," shade above the line; for "y <" or "y ≤," shade below.
Intersection of Inequalities
When dealing with systems of inequalities, like in our exercise, we look for the intersection of the solution regions of each inequality. Think of each inequality as creating its own "world" where certain points are valid solutions. The aim is to find where these "worlds" overlap. Here's how you can do that:
- First, graph each inequality on the same set of axes, just like we talked about in the previous section.
- Shade the relevant regions for each inequality. Remember, shading shows all potential solutions for that particular inequality.
- The intersection, or the area where both shadings overlap, represents the solution set for the system of inequalities. This is the set of points that satisfy both inequalities simultaneously.
- If there is no overlapping region, then there is no solution, meaning there's no point that satisfies all inequalities in the system.
Linear Inequality Solutions
Solving linear inequalities differs slightly from solving linear equations, mainly due to the inequality sign. The solution isn't just a single number but a set of numbers that makeup an entire region on a graph. Here's a helpful breakdown:
- Linear inequalities, just like linear equations, can be solved algebraically by isolating x or y. You rearrange to solve for "y" typically, as it makes plotting easier.
- When you multiply or divide by a negative number, always remember to flip the inequality sign.
- Once solved, use the graph to discover the practical set of solutions, which is usually visualized as an area rather than a specific point.
- In systems of inequalities, you're looking for all points that satisfy every inequality, and that’s where the intersection comes into play.
Other exercises in this chapter
Problem 29
Solve each system by the addition method. \(\left\\{\begin{array}{l}{3 x=4 y+1} \\ {3 y=1-4 x}\end{array}\right.\)
View solution Problem 29
Solve each system by the method of your choice. $$ \left\\{\begin{array}{l} {3 x^{2}+4 y^{2}=16} \\ {2 x^{2}-3 y^{2}=5} \end{array}\right. $$
View solution Problem 29
Solve each system for \((x, y, z)\) in terms of the nonzero constants \(a, b,\) and \(c .\) $$ \left\\{\begin{array}{r} {a x-b y-2 c z=21} \\ {a x+b y+c z=0} \\
View solution Problem 29
write the partial fraction decomposition of each rational expression. $$\frac{5 x^{2}-6 x+7}{(x-1)\left(x^{2}+1\right)}$$
View solution