Question

sketching the graph of a linear inequality in exercises 21–28, sketch the graph of the linear inequality. 21. $y \leq 3x$ 22. $y \geq 6x$ 23. $y \leq 2x - 3$ 24. $y \geq -x + 2$ 25. $x - y < 0$ 26. $x + y > 0$ 27. $y > -2x + 10$ 28. $y < 3x - 7$

Explanation:

Step1: Analyze the inequality type

The inequality \( y \leq 3x \) is a linear inequality in two variables. First, we consider the boundary line which is the equation \( y = 3x \). This is a straight line with a slope of \( 3 \) and a y - intercept of \( 0 \).

Step2: Determine the line type (dashed or solid)

Since the inequality is \( \leq \) (less than or equal to), the boundary line \( y = 3x \) should be a solid line. This is because the points on the line \( y = 3x \) are included in the solution set of \( y\leq3x \).

Step3: Test a point to find the shaded region

We can test a point that is not on the line. A common point to test is the origin \((0,0)\). Substitute \( x = 0 \) and \( y = 0 \) into the inequality \( y\leq3x \):
\( 0\leq3(0) \)
\( 0\leq0 \), which is a true statement. So, the origin \((0,0)\) lies in the solution region. Therefore, we shade the region that contains the origin (the region below or on the line \( y = 3x \)).

To sketch the graph:

1. Draw the solid line \( y = 3x \) (using a ruler to ensure a straight line, passing through points like \((0,0)\) and \((1,3)\) since for \( x = 1 \), \( y=3(1) = 3\)).
2. Shade the region that contains the origin (the region below the line \( y = 3x \)) because the test point \((0,0)\) satisfies the inequality \( y\leq3x \).

(The same process can be applied to other inequalities in the list:

• For \( y\geq6x \): Boundary line \( y = 6x \) (solid), test \((0,0)\): \( 0\geq0 \) is true, shade the region containing \((0,0)\) (above or on \( y = 6x \)).
• For \( y\leq2x - 3 \): Boundary line \( y=2x - 3 \) (solid), test \((0,0)\): \( 0\leq - 3\) is false, so shade the region not containing \((0,0)\) (below or on \( y = 2x-3\)).
• For \( y\geq - x + 2 \): Boundary line \( y=-x + 2 \) (solid), test \((0,0)\): \( 0\geq2\) is false, shade the region not containing \((0,0)\) (above or on \( y=-x + 2\)).
• For \( x - y\lt0\), rewrite as \( y>x \). Boundary line \( y = x \) (dashed, since the inequality is \( \lt\)), test \((0,0)\): \( 0>0\) is false, shade the region not containing \((0,0)\) (above \( y = x\)).
• For \( x + y>0\), rewrite as \( y>-x \). Boundary line \( y=-x \) (dashed), test \((0,0)\): \( 0>0\) is false, shade the region not containing \((0,0)\) (above \( y=-x\)).
• For \( y>-2x + 10 \): Boundary line \( y=-2x + 10 \) (dashed), test \((0,0)\): \( 0>-10\) is true, shade the region containing \((0,0)\) (above \( y=-2x + 10\)).
• For \( y\lt3x - 7 \): Boundary line \( y = 3x-7 \) (dashed), test \((0,0)\): \( 0\lt - 7\) is false, shade the region not containing \((0,0)\) (below \( y = 3x - 7\)).)

Answer:

To sketch \( y\leq3x \):

1. Draw the solid line \( y = 3x \) (passing through \((0,0)\) and \((1,3)\) etc.).
2. Shade the region below (and including) the line \( y = 3x \) (the region containing the origin \((0,0)\)).

(For other inequalities, follow the similar steps of identifying the boundary line (solid/dashed), testing a point, and shading the appropriate region.)