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Two-Variable Inequalities
Anna Askew
MAT 222 Week 2 Assignment
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Two-Variable Inequalities
Two-variable inequalities, just like the name suggests, are inequalities that have two variables and take the form of Ax + By > C. The inequality symbol can either be less than, greater than, greater than or equal to, or less than or equal to. Since we are dealing with equality, the graphs and solutions demonstrate a range of possible answers that would work in the given situation. Two-variable inequalities can be in navigation to guide the captains of the ships. This concept can be applied by ship operators to know the number of products they can ship during a trip.
All of the possibilities for the number of refrigerators and the number of TVs that will fit into an 18-wheeler are as shown in the graph below.
An equality that describes the shaded region can be derived using the data presented.
The diagram is showing the number of refrigerators on the x axis and the number of TVs on the y axis. There are two points on the graph, (110, 0) and (0, 330), so we can compute the slope of this line as follows (Dugopolski, 2013):
m = (Y1 – Y2) / (X1 – X2)
= (0 – 330) / (110 – 0)
= -330/110
= -3
This gradient will help form a linear inequality as follows:
Y1 – Y2 = m(X1 – X2)start with the point-slope form
Y1 – 0 = -3 (X1 – 110) substituting -3 for the gradient and (110, 0) for x and y
Y1 = -3X1 + 330use the distributive property
Y1 + 3X1 ≤ 330inert the less than or equal symbol after adding 3X1 to both sides
The graph has a solid line and not a dotted line; this is why the inequality symbol includes an equal to bar.
The next question asks if the truck will hold 71 refrigerators and 118 TVs.
This will form a point (71, 118) which is called a test point and is used to make sure the shading will fall on correct side of the line. To determine this we substitute it into the previously formed inequality as follows:
Y1 + 3X1 ≤ 330
118 + 3*71 ≤ 330
118 + 213 ≤ 330
331 ≤ 330this statement is false therefore the truck cannot hold 71 refrigerators and 118 TVs.
The next question is asking it the truck can hold 51 refrigerators and 176 TVs
This is another test point and its solution is similar to the previous one.
Y1 + 3X1 ≤ 330
176 + 3*51 ≤ 330
176 + 153 ≤ 330
329 ≤ 330the statement is true and so the truck can carry the load
The next problem requires a maximum to be found which can be calculated as follows:
Y1 + 3X1 ≤ 330
Y1 +3*60 ≤ 330replace X1 with 60
Y1 ≤ 150the maximum number of televisions is 110
The next problem also requires a maximum to be found which can be calculated as follows:
Y1 + 3X1 ≤ 330
200 + 3X1 ≤ 330replace Y1 with 200
3X1 ≤ 130subtract 200 from both sides
X1 ≤ 43.3333divide both sides by 3. Maximum number of refrigerators is 43
In summary, two-variable inequalities can be used to calculate how many items can fit in a given space. At times, one may have a storage or carriage space but lack the accuracy of determining how many of what items can fit. With the knowledge of two-variable inequalities they can be able to predict this. I have learnt that variable inequalities are very useful in many sectors. People who deal in transit business will find this concept very useful.
Reference
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.
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