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QUESTION IMAGE

graph the solution set of the following linear inequality $-2y \\leq 5x…

Question

graph the solution set of the following linear inequality
$-2y \leq 5x - 10$

answer
the line will be drawn once all required data is provided and will update whenever a value is updated. the regions will be added once the line is drawn.

choose the type of boundary line:
\\( \square \\) solid (---) \\( \quad \\) dashed (---)

enter two points on the boundary line:
( \\_\\_\\_\\_ , \\_\\_\\_\\_ ) ( \\_\\_\\_\\_ , \\_\\_\\_\\_ )

select the region you wish to be shaded:
\\( \square \\) a \\( \quad \\) \\( \square \\) b

Explanation:

Step1: Simplify the inequality

Start with the given inequality \(-2y \leq 5x - 10\). Divide both sides by \(-2\). Remember that when dividing or multiplying an inequality by a negative number, the direction of the inequality sign changes. So we get \(y \geq -\frac{5}{2}x + 5\).

Step2: Determine the boundary line type

Since the inequality is \(\geq\) (which includes equality), the boundary line should be solid.

Step3: Find two points on the boundary line

The boundary line is \(y = -\frac{5}{2}x + 5\) (the equation when we replace the inequality sign with an equal sign).

  • Let \(x = 0\): Then \(y = -\frac{5}{2}(0) + 5 = 5\). So one point is \((0, 5)\).
  • Let \(x = 2\): Then \(y = -\frac{5}{2}(2) + 5 = -5 + 5 = 0\). So another point is \((2, 0)\).

Step4: Determine the shaded region

We test a point not on the line, say \((0, 0)\) in the original inequality \(-2y \leq 5x - 10\). Substitute \(x = 0\) and \(y = 0\): \(-2(0) \leq 5(0) - 10\) which simplifies to \(0 \leq -10\), which is false. So we shade the region that does not include \((0, 0)\). Since \(y \geq -\frac{5}{2}x + 5\), we shade above the line. But for the purpose of the points and line type:

Answer:

  • Type of boundary line: Solid (—)
  • Two points on the boundary line: \((0, 5)\) and \((2, 0)\)
  • (For the shaded region, since when we tested \((0,0)\) it was false, we shade the region that satisfies \(y \geq -\frac{5}{2}x + 5\), which would be region B if the regions are set such that A is below and B is above, but based on the inequality \(y \geq\) the line, we shade above the line.)