Math: Systems of Inequalities
Solving and graphing systems of linear inequalities
Math: Systems of Inequalities
Solving and graphing systems of linear inequalities
Math - Grade 9-12
- 1
Solve the system and describe the solution set: y > 2x - 1 and y <= x + 3.
Graph both boundary lines first. Use a dashed line for > and a solid line for <=.
The solution set is all points that lie above the line y = 2x - 1 and on or below the line y = x + 3. These points satisfy both inequalities at the same time. - 2
Graph the system: x >= -2 and y < 4. Describe the region that is shaded.
The solution set is the region on or to the right of the vertical line x = -2 and below the horizontal line y = 4. The line x = -2 is solid, and the line y = 4 is dashed. - 3
Determine whether the point (2, 1) is a solution to the system: y < x and y >= -x + 2.
Substitute the coordinates into both inequalities.
Yes, the point (2, 1) is a solution. When x = 2 and y = 1, the inequality 1 < 2 is true and the inequality 1 >= 0 is also true. - 4
Determine whether the point (-1, 3) is a solution to the system: y <= 2x + 5 and y > -x + 1.
Yes, the point (-1, 3) is a solution. Substituting gives 3 <= 3, which is true, and 3 > 2, which is also true. - 5
Write the system of inequalities represented by this description: the region above the line y = -3x + 2 and on or below the line y = x - 1.
Above a line means y is greater than the line's equation.
The system is y > -3x + 2 and y <= x - 1. The first inequality is strict because the region is above the line, and the second includes the boundary because it is on or below the line. - 6
A school club sells notebooks for x dollars and pens for y dollars. The prices must satisfy x >= 1, y >= 0.5, and x + y <= 5. Describe what the solution set represents.
The solution set represents all possible notebook and pen prices that meet the minimum price rules and keep the total of one notebook and one pen at no more than 5 dollars. It is the region in the first quadrant on or above the minimum values and on or below the line x + y = 5. - 7
Solve the system: y < -2x + 6 and y > -2x + 1. Describe the region between the lines.
Both lines have the same slope, so they are parallel.
The solution set is all points between the parallel lines y = -2x + 6 and y = -2x + 1. Points must be below the first line and above the second line, so the shaded region is the strip between them, not including either boundary. - 8
Graph and describe the solution set for the system: y >= 3 and y < -x + 7.
The solution set is all points on or above the horizontal line y = 3 and below the line y = -x + 7. The overlap forms a region that includes the line y = 3 and excludes the line y = -x + 7. - 9
Find one point that satisfies the system y > x - 4 and y < 2x + 1. Explain why it works.
Choose an easy x-value, then test a y-value between the two expressions.
One possible point is (2, 0). It works because 0 > 2 - 4, so 0 > -2 is true, and 0 < 2(2) + 1, so 0 < 5 is also true. - 10
Is there a solution to the system y > x + 2 and y < x - 1. Explain your reasoning.
No, there is no solution. The line y = x + 2 is always 3 units above the line y = x - 1, so a point cannot be above the higher line and below the lower line at the same time. - 11
Write a system of two inequalities whose solution is the region in the first quadrant below the line y = -x + 6.
The first quadrant means both coordinates are nonnegative.
One correct system is x >= 0, y >= 0, and y <= -x + 6. These inequalities restrict the region to the first quadrant and keep all points on or below the line. - 12
A student must spend at least 4 hours on math, x, and science, y, combined, so x + y >= 4. The student also wants to spend no more than 6 hours total, so x + y <= 6, with x >= 0 and y >= 0. Describe the feasible region.
The feasible region is the set of all nonnegative points between the lines x + y = 4 and x + y = 6 in the first quadrant. It includes both boundary lines because the inequalities are inclusive. - 13
For the system y <= -x + 5 and y >= 2, find two ordered pairs that are solutions.
Pick points on or above y = 2 that also stay on or below y = -x + 5.
Two possible solutions are (1, 3) and (2, 2). For (1, 3), 3 <= 4 and 3 >= 2 are both true. For (2, 2), 2 <= 3 and 2 >= 2 are both true. - 14
Explain the difference between graphing y < 2x + 1 and y <= 2x + 1 in a system of inequalities.
For y < 2x + 1, the boundary line is dashed because points on the line are not included. For y <= 2x + 1, the boundary line is solid because points on the line are included in the solution set. - 15
A graph shows the overlap of these inequalities: x > 0, y > 0, and y < x + 2. Describe the solution set in words.
Strict inequalities use dashed boundaries and do not include the line.
The solution set is all points in the first quadrant that lie below the line y = x + 2. Because the inequalities are strict, the axes and the line are not included in the solution set.