Two variables - Speech or Presentation Example

Mathematics: Inequalities with two variables al affiliation Inequalities with two variables A linear inequality with two variables (x and y) is any inequality which can be written in the form Mx + Ny > P or Mx + Ny < P or Mx + Ny ≥ P or Mx + Ny ≤ P where,M, N and P are real numbers and M and N are not equal to zero. The problem being solved presents a two variable inequality application. The concept being examined (two variable inequalities) is important in real world setting since it enables us know all the number of ordered pairs of refrigerators and TVs that can be shipped in the 18-wheeler.

In writing the inequality that represents the region, we initially have to determine the equation of the solid boundary line. The region of interest is enclosed by the x and y axes and the boundary line. It is a gain very important to note that the boundary line is a solid line implying that the region of interest extends to the boundary line. Thus, we expect that our inequality will have the symbol ≤ or ≥ If our boundary line would be a dashed line instead, then we would have used the symbols < or > On the graph we have two coordinate points that will help in finding the equation of the line.

x-intercept (0,330) and y-intercept (110,0)From our knowledge of the equation of a line,G = ∆y / ∆xHence,G = (330 – 0) / (0 – 110)G = 330 / -110G = -3Now that we know the gradient of the boundary line and we have two coordinate points, we can find the equation of the line by choosing an arbitrary point(x, y) and substituting in the equation above,-3 = (y – 0) / (x – 110)-3 = y / (x – 110)Multiplying both sides by (x – 110) gives,-3(x – 110) = yRearranging y = -3x + 330After obtaining the equation of the boundary line, we will choose a test pint that will be true when substituted in the equation.

This test point must exist within the required region for the values to be true. We chose the origin (0, 0)Substituting in the equation of the boundary line y = -3x + 330 and comparing the left hand and right hand side of the equation, the inequality comes automatically. 0 ≠ -3(0) + 330 0 < 330Now that we remember that our boundary line was a solid line, the inequality representing the region becomes y ≤ -3x + 330The truck will not hold 71 refrigerators and 118 TVs because in substituting the values in the inequality, the result will not satisfy the inequalityy ≤ -3x + 330Substituting 71 for x and 118 for y,118 ≤ -3(71) + 330118 ≤ 117 Which is not true.

The truck will hold 51 refrigerators and 176 TVs because in substituting the values in the inequality, the result will satisfy the inequality y ≤ -3x + 330Substituting 51 for x and 176 for y, 176≤ -3(51) + 330176 ≤ 177Which is trueIn order to find the maximum number of TVs that will also be carried, we substitute 60 in the inequality to find the maximum number possible that will satisfy the inequality. hence,substituting 60 for x, y ≤ -3(60) + 330 y ≤ 150Thus the maximum number of TVs that will also be carried will be 150In order to find the maximum number of refrigerators that will also be carried, we substitute 200 in the inequality to find the maximum number possible that will satisfy the inequality.

Hence, substituting 200 for y, 200 ≤ -3x + 330Subtracting 330 from both sides of the inequality, -130 ≤ -3xDividing both sides of the inequality by -3 reverses the sign of the inequality as follows (-130 / -3x) ≥ xSince 130 is not divisible by 3, we have to get a number closest to 130 that is divisible by 3. The number is 126. In order to obtain 126 on the left hand side of the inequality, we must adjust the number of TVs. We thus introduce a linear equationt – 330 = -126t = 204The number of TVs to be carried will be 204.

Going back to our previous substitution, 204 ≤ -3x + 330-126 ≤ -3x42 ≤ xTherefore the maximum number of TVs to be carried will be 42Reference Kaufmann, J. E., & Schwitters, K. (2000). Intermediate algebra (6th ed.). Pacific Grove, CA: Brooks/Cole.