Sketching the Solution Set of a Linear Inequality



Analytic-Geometry:

One may extend the one dimensional analytical-geometry concepts in visualization of other algebraic elements,

[pic]

such as equations. For example, the above figure is a graph (i.e., a picture) of equation Y = X + 1, which is a straight line.

Co-ordinates: To find the co-ordinates of a graph the axis have to be identified. The vertical axis is known as the 'y' axis, and the horizontal axis is known as the 'x' axis. The position of a point on a graph is defined by its 'x' co-ordinate first, and its 'y' co-ordinate second. The point A is at x = 3, and y = 2, or A(3,2). When there are more than one point on a graph they are connected by a straight line or a curved line.

In this overview, we will start with graphing straight lines, and then progress to other graphs. The only major difference, really, is in how many points you need to plot in order to draw a good graph. But those increased numbers of points will vary with the "interesting" issues related to the various types of graphs.

Before we get started, though, let me say this: You should do NEAT graphs, which means that you should be using a ruler. If you don't have a ruler, go get one. Now. It will help immensely, and you can get major "brownie points" from your instructor. And, no, using graph paper does NOT excuse you from using a ruler.

Suppose you have "y = 2x + 3". Since this has just "x", as opposed to "x2" or "1/x", this graphs as just a plain straight line.

Then you will pick values for x, plug them into the equation, and solve for the corresponding values of y.  Don't forget to pick negatives for x; using only positive numbers can be misleading later on, so it's a bad habit to get in to now. Also, try to plot at least three points. It's just safer that way: if you mess up on one point, you'll know, because it's dot won't line up with the others. This is what this looks like:

|X |Create a chart to hold x and y values from your line.  The x-values may be any values you wish. |

|Y |Substitute the x-values into the equation to determine the y-values. |

| |Plot the (x, y) coordinates to graph the line. |

|-3/2 | |

| 0 | |

| | |

|-1 | |

| 1 | |

| | |

|0 | |

| 3 | |

| | |

|1 | |

| 5 | |

| | |

|2 | |

| 7 | |

| | |

|While charts often contain more than 2 entries, only two entries are needed to determine a straight line.  A third point should be used |

|to "check" that an error was not made while computing the first two points. |

[pic]

[pic]

Graph of Y = 2X + 3

|Horizontal Line |Vertical Line |

|y = 3 (or any number) |x = -2 (or any number) |

|Lines that are horizontal have a slope of zero.  They have "run", but no |Lines that are vertical have no slope (it does not exist).  They have |

|"rise".   The rise/run formula for slope always yields zero since rise = |"rise", but no "run".  The rise/run formula for slope always has a zero |

|0. |denominator and is undefined. |

|y = mx + b |These lines are described by what is happening to their x-coordinates.  |

|y = 0x + 3 |In this example, the x-coordinates are always equal to -2. |

|y = 3 |Below is an illustration of a vertical line x = c: |

|This equation also describes what is happening to the y-coordinates on |[pic] |

|the line.  In this case, they are always 3. | |

|Below is an illustration of a horizontal line y = c: | |

|[pic] | |

Homework I: Consider the graph at the right containing three lines. Which of the following equations matches the lines A, B, and C?

|y = 4 + (1/3) x |[pic] |

|y = 6 | |

|y = -1/3 + 4 x | |

|y = x + 6 | |

|y = 4 - (1/3) x | |

|y = (3/2)x + 2 | |

Homework II: Consider the graph containing two lines, which of the following equations matches the lines A and B.

|y = 4 - 2 x |[pic] |

|y = (2/3)x + 14 | |

|y = 1/2 x + 4 | |

|y = 4 + 2 x | |

|y = 14 - (3/2) x | |

|y = 14 - (2/3) x | |

Sketching the Solution Set of a Linear Inequality

To sketch the region represented by a linear inequality in two variables:

A. Sketch the straight line obtained by replacing the inequality with equality.

B. Choose a test point not on the line ((0,0) is a good choice if the line does not pass through the origin, and if the line does pass through the origin a point on one of the axes would be a good choice).

C. If the test point satisfies the inequality, then the set of solutions is the entire region on the same side of the line as the test point. Otherwise it is the region on the other side of the line. In either case, shade out the side that does not contain the solutions, leaving the solution region showing.

Example

To sketch the linear inequality

3x - 4y ≤ 12,

first sketch the line 3x - 4y = 12.

[pic]

Next, choose the origin (0, 0) as the test point (since it is not on the line). Substituting x=0, y=0 in the inequality gives

3(0) - 4(0) ≤ 12.

Since this is a true statement, (0, 0) is in the solution set, so the solution set consists of all points on the same side as (0, 0). This region is left unshaded, while the (grey) shaded region is blocked out.

[pic]

Graphing Lines: Take for example the line 2 x – 3 y – 6 = 0. Now let us draw this line using our Coordinate Geometry. Remember from our high school that we often find where the line cuts the X and Y axes. To do this let x = 0 and y = 0 respectively. We can find and plot 2 points on the line, they are (0, 2) and (3, 0).

 

[pic]

All the points on the above line are represented by the equation 2 x -3 y – 6 = 0.

Inequalities and Feasible Region

The feasible region determined by a collection of linear inequalities is the collection of points that satisfy all of the inequalities.

To sketch the feasible region determined by a collection of linear inequalities in two variables: Sketch the regions represented by each inequality on the same graph, remembering to shade the parts of the plane that you do not want. What is unshaded when you are done is the feasible region.

Example

The feasible region for the following collection of inequalities is the unshaded region shown below (including its boundary).

3x - 4y ≤ 12, x + 2y ≥ 4, x ≥ 1, y ≥ 0.

[pic]

A three dimensional feasible region:

Consider the following 3-dimensional feasible region:

X1 + X2 + X3 ≤ 10

3X1 + X3 ≤ 24

X1, X2, and X3 ≥ 0.

The feasible region is shown in the following figure.

[pic]

Note that, the feasible region is defined by a set of five constraints (which includes the three signed variables). The feasible region has six vertices as shown above, and table below:

|X1 = 8     |X1 = 8     |X1 = 0     |X1 = 0     |X1 = 7     |X1 = 0     |

|X2 = 0 |X2 = 2 |X2 = 10 |X2 = 0 |X2 = 0 |X2 = 0 |

|X3 = 0 |X3 = 0 |X3 = 0 |X3 = 10 |X3 = 3 |X3 = 0 |

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