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x=-4|y|+3
Find four points contained in the inverse. Express your values as an integer or simplified fraction.

x=4y+3x=-4|y|+3\newlineFind four points contained in the inverse. Express your values as an integer or simplified fraction.

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Q. x=4y+3x=-4|y|+3\newlineFind four points contained in the inverse. Express your values as an integer or simplified fraction.
  1. Interchanging roles of xx and yy: To find the inverse of the given relation, we need to solve for yy in terms of xx. This means we will interchange the roles of xx and yy and then solve for yy. The original relation is x=4y+3x = -4|y| + 3.
  2. Expressing relation with interchanged variables: First, we express the relation with yy and xx interchanged: y=4x+3y = -4|x| + 3.
  3. Considering absolute value: Now, we need to consider the absolute value. Since |x|\
  4. Case \(1: xx is non-negative: Case 11: If xx is non-negative (x0x \geq 0), then x=x|x| = x, and the equation becomes y=4x+3y = -4x + 3.
  5. Finding points for Case 11: From Case 11, we can find two points by choosing non-negative values for xx and calculating the corresponding yy values. Let's choose x=0x = 0 and x=1x = 1.
  6. Case 22: xx is negative: For x=0x = 0: y=4(0)+3=3y = -4(0) + 3 = 3. So, one point is (0,3)(0, 3).
  7. Finding points for Case 22: For x=1x = 1: y=4(1)+3=1y = -4(1) + 3 = -1. So, another point is (1,1)(1, -1).
  8. Finding points for Case 22: For x=1x = 1: y=4(1)+3=1y = -4(1) + 3 = -1. So, another point is (1,1)(1, -1).Case 22: If xx is negative (x < 0), then x=x|x| = -x, and the equation becomes y=4(x)+3=4x+3y = -4(-x) + 3 = 4x + 3.
  9. Finding points for Case 22: For x=1x = 1: y=4(1)+3=1y = -4(1) + 3 = -1. So, another point is (1,1)(1, -1).Case 22: If xx is negative (x < 0), then x=x|x| = -x, and the equation becomes y=4(x)+3=4x+3y = -4(-x) + 3 = 4x + 3.From Case 22, we can find two more points by choosing negative values for xx and calculating the corresponding yy values. Let's choose x=1x = -1 and y=4(1)+3=1y = -4(1) + 3 = -100.
  10. Finding points for Case 22: For x=1x = 1: y=4(1)+3=1y = -4(1) + 3 = -1. So, another point is (1,1)(1, -1).Case 22: If xx is negative (x < 0), then x=x|x| = -x, and the equation becomes y=4(x)+3=4x+3y = -4(-x) + 3 = 4x + 3.From Case 22, we can find two more points by choosing negative values for xx and calculating the corresponding yy values. Let's choose x=1x = -1 and y=4(1)+3=1y = -4(1) + 3 = -100.For x=1x = -1: y=4(1)+3=1y = -4(1) + 3 = -122. So, another point is y=4(1)+3=1y = -4(1) + 3 = -133.
  11. Finding points for Case 22: For x=1x = 1: y=4(1)+3=1y = -4(1) + 3 = -1. So, another point is (1,1)(1, -1).Case 22: If xx is negative (x < 0), then x=x|x| = -x, and the equation becomes y=4(x)+3=4x+3y = -4(-x) + 3 = 4x + 3.From Case 22, we can find two more points by choosing negative values for xx and calculating the corresponding yy values. Let's choose x=1x = -1 and y=4(1)+3=1y = -4(1) + 3 = -100.For x=1x = -1: y=4(1)+3=1y = -4(1) + 3 = -122. So, another point is y=4(1)+3=1y = -4(1) + 3 = -133.For y=4(1)+3=1y = -4(1) + 3 = -100: y=4(1)+3=1y = -4(1) + 3 = -155. So, the last point is y=4(1)+3=1y = -4(1) + 3 = -166.

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