Linear inequalities are used to represent the relationship between two variables where one is greater than or less than the other. Graphing these inequalities can be useful in understanding their solution set and providing a visual representation of the relationship between the variables.
The linear inequality 2x – 3y < 12 can be graphed by first finding its boundary line, which is the line that represents the equation 2x – 3y = 12. This line separates the plane into two regions: one where 2x – 3y is greater than 12, and one where it is less than 12. The region where 2x – 3y is less than 12 is the solution set of the inequality.
To find the boundary line, we can rearrange the equation 2x – 3y = 12 to solve for y:
2x – 3y = 12 -3y = -2x + 12 y = (2/3)x - 4
This is a line with slope 2/3 and y-intercept -4. We can plot this line by finding two points on it, or by using its slope and y-intercept to draw a straight line.
To find two points on the line, we can choose any two values of x and plug them into the equation y = (2/3)x - 4 to find the corresponding values of y. For example, if we set x = 0, we get y = -4, so one point on the line is (0, -4). If we set x = 3, we get y = -2, so another point on the line is (3, -2). We can plot these points and draw a straight line passing through them to represent the boundary line.
Alternatively, we can use the slope-intercept form of the equation y = mx + b, where m is the slope and b is the y-intercept, to draw the line. In this case, the slope is 2/3 and the y-intercept is -4, so we can start at the y-intercept (0, -4) and use the slope to find another point on the line. The slope tells us that for every increase of 3 in x, y increases by 2, so we can add 3 to x and 2 to y to get another point on the line: (3, -2). We can plot these two points and draw a straight line passing through them to represent the boundary line.
Once we have the boundary line, we need to determine which side of it represents the solution set of the inequality 2x – 3y < 12. To do this, we can choose a test point in one of the regions and substitute its coordinates into the inequality. If the inequality is true, then that region is part of the solution set; if the inequality is false, then that region is not part of the solution set.
A convenient test point to use is (0, 0), which is the origin. We can substitute x = 0 and y = 0 into the inequality 2x – 3y < 12 to get:
2(0) – 3(0) < 12 0 < 12
Since this is true, the region containing the origin is part of the solution set. To determine which side of the boundary line contains the origin, we can use the fact that the side containing the origin is the same side as the side where the inequality is satisfied. In this case, the inequality is satisfied when 2x – 3y is less than 12, so the region containing the origin is the one where 2x – 3y is less than
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