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Look at the system of inequalities.\newliney13x+3y \leq -\frac{1}{3}x + 3\newlinex0x \geq 0\newliney0y \geq 0\newlineThe solution set is the triangular region where all the inequalities are true.\newlineWhat are the vertices of that triangular region?\newline(____,____)\newline(____,____)\newline(____,____)

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Q. Look at the system of inequalities.\newliney13x+3y \leq -\frac{1}{3}x + 3\newlinex0x \geq 0\newliney0y \geq 0\newlineThe solution set is the triangular region where all the inequalities are true.\newlineWhat are the vertices of that triangular region?\newline(____,____)\newline(____,____)\newline(____,____)
  1. Find Intersection with X-Axis: First, let's find the intersection of y=13x+3y = -\frac{1}{3}x + 3 and the x-axis (y=0y = 0).\newlineSet yy to 00 and solve for xx: 0=13x+30 = -\frac{1}{3}x + 3.\newlineMultiply both sides by 33: 0=x+90 = -x + 9.\newlineAdd xx to both sides: x=9x = 9.\newlineSo, the intersection with the x-axis is at y=0y = 000.
  2. Find Intersection with Y-Axis: Next, find the intersection of y=13x+3y = -\frac{1}{3}x + 3 and the y-axis (x=0x = 0).\newlineSet xx to 00 and solve for yy: y=13(0)+3y = -\frac{1}{3}(0) + 3.\newlineSimplify: y=3y = 3.\newlineSo, the intersection with the y-axis is at (0,3)(0, 3).
  3. Find Origin Intersection: The third vertex is the intersection of the lines x=0x = 0 and y=0y = 0, which is the origin (0,0)(0, 0).
  4. Identify All Three Vertices: Now we have all three vertices of the triangular region: (9,0)(9, 0), (0,3)(0, 3), and (0,0)(0, 0).

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