Two Variable Inequalities
Cory R. Hambley
Math 222: Intermediate Algebra
Instructor Laura Cella
October 19, 2014

Two Variable Inequalities

Two variable inequalities have several practical applications that make them valuable in the real world. Some good examples would be price comparisons and logistics and shipping. By comparing two variables, dependent and independent, one can examine the relationship and effects that these variables have on each other in given scenarios. There are also graphs that correspond with the solutions to these inequalities. The graphs are important because they give the best representation of the answer range for the inequality that is given. This week’s assignment is problem 68 on page 539 of our text (Dugopolski, 2012). Its corresponding graph shows all of the possibilities for the number of TVs and refrigerators that can fit into an 18-wheeler truck for shipping. An 18-wheeler can fit a maximum of 110 refrigerators and zero TVs or a maximum of zero refrigerators and 330 TVs.

On this graph the refrigerators are represented by the x and the TVs are represented by the y. The two point on the graph, (0, 330) and (110, 0) represent two points. With these points we can determine the slope of the line and write a point-slope equation and then transform it into a linear inequality. For the first part of this problem we will write an inequality to describe this region.
To find the slope: m = y₁ - y₂ = 330 – 0 = -330 = -3 x₁ - x₂ 0 – 110 110

y - y₁ = m(x - x₁) Point-Slope form

y – 330 = -3 (x – 0) Substituting point (330. 0) and m with previously found slope. y = -3x + 330 Add 300 to both sides

y + 3x < 330 Add 3x to both sides. Change the equal sign to a “less than or equal to” sign. The graph of this inequality will be a solid line rather than a dashed line. This means it is included in the range of the solutions. This is true because the line itself is a solution in the range of solutions. A dashed line would NOT be included in the range of solutions.
Let’s add a test point to see if this holds true ad to check our range of solutions. Can the 18-wheeler hold 71 refrigerators and 118 TVs? The test point will be (71, 118). It will be used to determine whether the correct side of the graph is shaded and in order to check this we have to substitute it into the inequality to ensure that it remains true. y + 3x < 330 118 + 3(71) < 330 Substitute the test point values. 118 + 213 < 330 331 < 330
This inequality is not true, making the statement untrue. So, one TV will not fit into the 18 wheeler.
Next, will the truck hold 51 refrigerators and 176 TVs? My test point will be (51, 176). Again, it will be used to determine whether the correct side of the graph is shaded and in order to check this we have to substitute it into the inequality to ensure that it remains true. y + 3x < 330 176 + 3(51) < 330 Substitute the test point (x, y) values. 176 + 153 < 330 329 < 330
This inequality is true. We can still fit everything in and have room to spare.
The next scenario is a Buy More Store Max Order. If they buy, at most, 60 refrigerators, how many TVs can they fit in the truck? In order to find the maximum number of TVs, 60 must be substituted into the linear equality for x and then y must be solved for. So: y + 3x < 330 y + 3(60) < 330 Substitute the given value and multiply. y + 180 < 330 Subtract 180 from both