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Chapter 11: Problem 712

Graph the system \(\mathrm{x} \geq 4\) and \(2 x \leq 18\)

Short Answer

Expert verified
In conclusion, the solution to the given system is all values of x that satisfy \(4\leq x\leq9\). This can be represented on a number line graph with a shaded region that starts from a solid circle at x = 4 and ends at a solid circle at x = 9.

Step by step solution

01

Solve the inequalities for x.

First, let's solve each inequality equation for x. 1. \(\mathrm{x} \geq 4\): This inequality is already solved for x. 2. \(2x \leq 18\): Divide both sides by 2 to solve for x: \(x \leq \frac{18}{2}\), giving us \(x \leq 9\). Now we have two inequalities: \(\mathrm{x} \geq 4\) and \(\mathrm{x} \leq 9\).
02

Graph the solution region of each inequality.

Now that we have both inequalities solved for x, we can graph each inequality's solution region on a number line: 1. For the inequality \(\mathrm{x} \geq 4\), draw a solid circle on the number line at x = 4 (since x can be equal to 4) and shade the region to the right of x = 4 (since x can be greater than 4). 2. For the inequality \(\mathrm{x} \leq 9\), draw a solid circle on the number line at x = 9 (since x can be equal to 9) and shade the region to the left of x = 9 (since x can be less than 9).
03

Find the intersection of both solution regions.

Now, we need to find the intersection of both solution regions. To do this, look for the region where both shaded regions overlap on the number line: The shaded region starts at x = 4 (inclusive) and goes up to x = 9 (inclusive). So the intersection of both solution regions is \(4\leq x\leq9\). In conclusion, the solution to the given system is all values of x that satisfy \(4\leq x\leq9\). This can be represented on a number line graph with a shaded region that starts from a solid circle at x = 4 and ends at a solid circle at x = 9.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

System of Inequalities
Understanding a system of inequalities is crucial when dealing with multiple constraints at the same time. A system of inequalities refers to two or more inequalities that work together to restrict the range of possible solutions. In the given exercise, we have a system with two parts:

1. \(x \geq 4\) 2. \(2x \leq 18\)

Here, the challenge is to find all the values of \(x\) that satisfy both conditions simultaneously. This forms the basis for finding the intersection or the common solution set. It's similar to a puzzle where all pieces need to fit together to complete the picture.
Number Line Representation
Graphical representation is a powerful tool for visualizing mathematical concepts, and this holds true for inequalities as well. A number line is a simple yet effective way to illustrate the solutions of inequalities. For each inequality in the system, we draw a line with a starting point that is either open or closed, depending on whether the inequality is strict ('greater than' or 'less than') or inclusive ('greater than or equal to' or 'less than or equal to').

For \(x \geq 4\), we put a solid dot on 4 and shade rightward to indicate all numbers greater than or equal to 4. Similarly, for \(x \leq 9\), a solid dot on 9 and shading leftward signifies all numbers less than or equal to 9. This visual helps students identify the solution sets easily.
Solving Inequalities
Solving inequalities is somewhat analogous to solving equations, but with a twist. You perform similar operations to isolate the variable on one side. However, you must remember that multiplying or dividing by a negative number reverses the direction of the inequality.

In the exercise, we solved \(2x \leq 18\) by dividing both sides by 2, maintaining the direction of the inequality because we divided by a positive number. As a result, we got \(x \leq 9\). It's essential to follow these rules to arrive at the correct solution set.
Intersection of Solution Regions
The final step in handling a system of inequalities is to find the common solution, or the intersection, of all individual inequalities. On a number line, this is the area where all shaded regions overlap.

In our example, the overlapping region is from the solid dot at 4 to the solid dot at 9, inclusive of both ends. It means that the solution for the system is all real numbers between 4 and 9, including 4 and 9 themselves. This intersection is the 'sweet spot' where all the conditions of the system are met. Identifying this region correctly is critical for solving the system.

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Most popular questions from this chapter

Find a basic feasible solution to: $$ \begin{aligned} &2 \mathrm{x}_{1}+\mathrm{x}_{2}+2 \mathrm{x}_{3}=4 \\ &3 \mathrm{x}_{1}+3 \mathrm{x}_{2}+\mathrm{x}_{3}=3 \\ &\mathrm{x}_{1} \geq 0, \mathrm{x}_{2} \geq 0, \mathrm{x}_{2} \geq 0 \end{aligned} $$

A businessman needs 5 cabinets, 12 desks, and 18 shelves cleaned out. He has two part time employees Sue and Janet. Sue can clean one cabinet, three desks and three shelves in one day, while Janet can clean one cabinet, two desks and 6 shelves in one day. Sue is paid \(\$ 25\) a day, and Janet is paid \(\$ 22\) a day. In order to minimize the cost how many days should Sue and Janet be employed?

Minimize \(\quad \mathrm{x}_{1}+\mathrm{x}_{2}\) Subject to: \(\begin{aligned} & \mathrm{x}_{1}+2 \mathrm{x}_{2} \leq 4 \\ & \mathrm{x}_{2} \leq 1 \\ & \mathrm{x}_{1}, \mathrm{x}_{2} \geq 0 \end{aligned}\) Find a basic feasible solution to the above problem, starting from a b.f.s with xi and in the basis.

Consider the following minimization problem: $$ \min z=2 x_{1}+x_{2}-x_{3} $$ subject to: $$ \begin{aligned} &\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3} \leq 3 \\ &\mathrm{x}_{2}+\mathrm{x}_{3} \geq 2 \\ &\mathrm{x}_{1}+\mathrm{x}_{3}=1 \\ &\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3} \geq 0 \end{aligned} $$ Find an initial basic feasible solution.

Find nonnegative numbers \(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}\) which maximize $$ 3 \mathrm{x}_{1}+\mathrm{x}_{2}+9 \mathrm{x}_{3}-9 \mathrm{x}_{4} $$ and satisfy the conditions $$ \begin{aligned} &\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}-5 \mathrm{x}_{4}=4 \\ &\mathrm{x}_{1}-\mathrm{x}_{2}+3 \mathrm{x}_{3}+\mathrm{x}_{4}=0 \end{aligned} $$
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