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Two-Variable Inequalities
Name:
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From 68 on page 539 of Elementary and Intermediate Algebra:
Graph one shows the possibilities on the number of refrigerators and TVs that can fit into the 18
wheeler. The truck can hold zero TVs but hold a maximum of 110 refrigerators at a single take
or hold zero refrigerators and 330 TV’s at a single take.
(x,y)
(0,330)
(x,y)
(110,0)
Graph 1
From graph1: x and y represent refrigerators and TVs respective. From the two values, the line
gradient can be used to determine the linear inequality.
The linear inequality can be estimated as
a) An inequality to describe this region.
x = refrigerators
y = TV
Substitute point (0, 330)
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The inequality will be
y+3x <= 330;
The second inequality will be
b) Can the wheeler accommodate 71 refrigerators and 118 TVs?
Point (71,118) will the test point for checking if the solution is affirmative (if it will fall on
the line right side region), substituting (71,118) to the equation will be…
3x71 + 118 <= 330
331 <= 330
because the answer is less than 330, this will be NO
Using another test point (51,176), the third inequality will be
c) Can the wheeler hold 176 TVs and 51 refrigerators at the same time?
176 +3 x 51 <= 330
329 <= 330
the answer is YES sine 329 is less than 330
d) The second problem, maximum and minimum values will be estimated by;
First estimating the maximum number t solve for the two unknowns.
This will be
𝑦 + 3𝑥 ≤ 330
𝑦 + 3(60) ≤ 330
𝑦 + 180 ≤ 330
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The value of y will be less than 150
𝑦 ≤ 150
Conclusion is that, rather than the 60 refrigerators, extra 150 TVs can be fitted.
This can graphically be shown by sketching a vertical solid line rather than dashed line at
x= 60 and a horizontal line at y=150. Joining the two lines, a rectangular shape will be
formed. From this shaded rectangular form, we will have only one possible solution for the
amount that will fit in the 18-wheeler.
e) If Burbank Buy More decides to have a television sale and include 200 TVs, The maximum
number of refrigerators to be delivered at a given trip will be as follows
𝑦 + 3𝑥 ≤ 330;
200 + 3𝑥 ≤ 330;
3𝑥 ≤ 130;
𝑥 ≤ 43;
This can be concluded that, 43 maximum refrigerators can be delivered together with the 200
TV’s. Sketching a line similar to the previous, the horizontal line will be =200. The triangular
shape shaded above the line to will be used to find the number that can fit together with
the minimum number of refrigerators.
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,
(0,150)
y=200
Graph2
From graph 2, the two variable inequalities are ideal in solving day to day problems. We
would have estimated the inequality then draw a graph to show the expected solution if there was
more than one point in the line. According to Peterson (2004), using linear inequality method, is
of important value to a firm or a society in making huge returns.
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Reference
Peterson, J. C. (2004). Technical mathematics. Clifton Park, NY: Thomson/Delmar Learning.