Question

Which of the following is a point in the feasible region determined by the linear inequations \(2 x+3 y \leq 6\) and \(3 x-2 y \leq 16 ?\) (1) \((4,-3)\) (2) \((-2,4)\) (3) \((3,-2)\) (4) \((3,-4)\)

Short answer

(1) (4, -3) (2) (-2, 4) (3) (3, -2) (4) (3, -4) Answer: (3) (3, -2)

Step-by-step solution

Graph the inequalities on a coordinate plane

First, we need to rewrite the inequalities in slope-intercept form: 1. \(2x + 3y \leq 6\) Divide by 3: \(y \leq -\frac{2}{3}x + 2\) 2. \(3x - 2y \leq 16\) Divide by -2: \(y \geq \frac{3}{2}x - 8\) Now, we can graph these inequalities on a coordinate plane. Remember that for \(y \leq -\frac{2}{3}x + 2\), we shade below the line, and for \(y \geq \frac{3}{2}x - 8\), we shade above the line. The feasible region is the area where both inequalities are true (intersection of both shaded areas).

Check if each given point is in the feasible region

Now, we insert each of the given points into both inequalities to see if they are true: 1. Point \((4, -3)\): \(2(4) + 3(-3) = 8 - 9 = -1\) which is less than 6 - inequality 1 is true. \(3(4) - 2(-3) = 12 + 6 = 18\) which is greater than 16 - inequality 2 is not true. 2. Point \((-2, 4)\): \(2(-2) + 3(4) = -4 + 12 = 8\) which is greater than 6 - inequality 1 is not true. \(3(-2) - 2(4) = -6 - 8 = -14\) which is less than 16 - inequality 2 is true. 3. Point \((3, -2)\): \(2(3) + 3(-2) = 6 - 6 = 0\) which is less than 6 - inequality 1 is true. \(3(3) - 2(-2) = 9 + 4 = 13\) which is less than 16 - inequality 2 is true. 4. Point \((3, -4)\): \(2(3) + 3(-4) = 6 - 12 = -6\) which is less than 6 - inequality 1 is true. \(3(3) - 2(-4) = 9 + 8 = 17\) which is greater than 16 - inequality 2 is not true.

Identify the point in the feasible region

After checking all given points, we find that only \((3, -2)\) satisfies both inequalities. So, the correct answer is: (3) \((3,-2)\)

Key concepts

Graphing Linear Inequalities

When working with linear inequalities on a coordinate plane, it's essential to understand that graphing these inequalities will help visualize the solutions. For instance, if we take the inequality \(2x + 3y \leq 6\), to graph this, we would first convert it into slope-intercept form, which tells us where to plot the y-intercept and how to draw the line based on the slope.

Once in slope-intercept form, the inequality becomes \(y \leq -\frac{2}{3}x + 2\). The line \(y = -\frac{2}{3}x + 2\) is drawn on the graph, and because the inequality is \('less than or equal to'\), we shade below the line. This shading represents all the possible solutions to the inequality. The same process is done for any additional inequalities. The feasible region, where the solutions to all inequalities overlap, is the key area of interest.

Slope-Intercept Form

Slope-intercept form is a way to write linear equations such that they take the form \(y = mx + b\), where \(m\) is the slope of the line and \(b\) is the y-intercept. This form is extremely helpful for graphing because it gives you a starting point (the y-intercept) and a way to determine the direction and steepness of the line (the slope).

In our original inequalities, we converted them to slope-intercept form to graph them: \(y \leq -\frac{2}{3}x + 2\) and \(y \geq \frac{3}{2}x - 8\). These forms show that the lines cross the y-axis at 2 and -8, respectively, with slopes of \(-\frac{2}{3}\) and \(\frac{3}{2}\). This insight is crucial for accurately drawing the lines on a graph.

Coordinate Plane

The coordinate plane is a two-dimensional surface where each point is determined by a pair of numbers, known as coordinates. These coordinates are measured along two perpendicular axes, the x-axis (horizontal) and the y-axis (vertical).

The coordinate plane is fundamental in graphing linear inequalities because it allows you to plot the lines and regions that represent the solutions. For example, the point \((3, -2)\) is located by moving 3 units to the right of the origin and 2 units down. Understanding how to navigate this plane is vital to solving systems of inequalities, as seen in the original exercise.

Systems of Inequalities

A system of inequalities consists of multiple inequalities that are considered together. The solution to a system of inequalities is the set of points that satisfy all inequalities in the system simultaneously. In the context of our problem, we graphed the inequalities \(y \leq -\frac{2}{3}x + 2\) and \(y \geq \frac{3}{2}x - 8\) on the coordinate plane and looked for the intersection of the regions that complied with both.

The feasible region is effectively the 'common ground' among all inequalities. Each point in the feasible region, like \((3, -2)\) in our exercise, is a solution to the system. This process of checking each point against the inequalities allowed us to identify the single point that was the solution to the system.