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Algebra

Algebra Help: Graphing Linear Inequalities And Systems

Review real example questions for Graphing Linear Inequalities And Systems in Algebra.

Question 1

Is point (2,1)(2,1)(2,1) in the solution region of the inequality y<x−2y < x - 2y<x−2?

  1. Yes, because 1<01 < 01<0
  2. Yes, because 1<41 < 41<4
  3. No, because 1≮01 \not< 01<0
  4. No, because 1≮−41 \not< -41<−4
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. The solution to a linear inequality is an entire region (a half-plane), not just a single point: every point in the shaded region makes the inequality true! This is different from linear equations, which have just one solution point where the lines cross. To check if point (2, 1) is in the solution region of y < x - 2, we substitute x = 2 and y = 1 into the inequality: 1 < 2 - 2, which simplifies to 1 < 0. Since 1 is NOT less than 0, this statement is false, so the point is not in the solution region. Choice C correctly identifies that the point is not in the solution region because 1 ≮ 0 (1 is not less than 0). Great work! Choices A and B incorrectly claim the point is in the solution (with A making the false claim that 1 < 0), while choice D uses the wrong calculation comparing 1 to -4 instead of 0. For shading direction with y inequalities: y > [line] means 'y is greater than the line' = shade above (higher y-values). y < [line] means 'y is less than the line' = shade below (lower y-values). Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 2

Graph the solution set to the system {y≥2xy≤−x+6\begin{cases} y\ge 2x \\ y\le -x+6 \end{cases}{y≥2xy≤−x+6​ Which description is correct?

  1. The region below y=2xy=2xy=2x and above y=−x+6y=-x+6y=−x+6.
  2. The region above y=2xy=2xy=2x and above y=−x+6y=-x+6y=−x+6.
  3. The region above y=2xy=2xy=2x and on or below y=−x+6y=-x+6y=−x+6.
  4. The region on or below y=2xy=2xy=2x and on or below y=−x+6y=-x+6y=−x+6.
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. A system of linear inequalities has a solution region that's the intersection (overlap) of all the individual half-planes: you graph each inequality, and where all the shaded regions overlap is where all the inequalities are satisfied at once. That intersection is your feasible region! For the system y ≥ 2x and y ≤ -x + 6, we need points that satisfy both conditions. The first inequality y ≥ 2x includes all points on or above the line y = 2x. The second inequality y ≤ -x + 6 includes all points on or below the line y = -x + 6. The solution region is where these overlap: above (or on) y = 2x AND below (or on) y = -x + 6. Choice C correctly describes the solution region as above y = 2x and on or below y = -x + 6, which captures the intersection of both half-planes. Great work! Choice A incorrectly says 'below y = 2x', choice B says 'above' for both (missing the intersection), and choice D says 'below' for both (also missing the correct regions). For systems, think of it like finding what's allowed: each inequality restricts the plane, and the solution is where ALL the restrictions are met simultaneously. The key is to carefully track which side of each boundary line is shaded, then find where all shadings overlap!

Question 3

Describe the solution region for the system of inequalities {x≥0y≥0y≤−2x+6\begin{cases}x\ge 0\\y\ge 0\\y\le -2x+6\end{cases}⎩⎨⎧​x≥0y≥0y≤−2x+6​

  1. All points in the first quadrant that are on or below the line y=−2x+6y=-2x+6y=−2x+6, including the boundary.
  2. All points with x≤0x\le 0x≤0 and y≤0y\le 0y≤0 that are below the line y=−2x+6y=-2x+6y=−2x+6.
  3. All points in Quadrant II on or above the line y=−2x+6y=-2x+6y=−2x+6.
  4. All points in the first quadrant that are on or above the line y=−2x+6y=-2x+6y=−2x+6, excluding the boundary.
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. A system of linear inequalities has a solution region that's the intersection (overlap) of all the individual half-planes: you graph each inequality, and where all the shaded regions overlap is where all the inequalities are satisfied at once. That intersection is your feasible region! For this system, we have three constraints: x ≥ 0 (points on or to the right of the y-axis), y ≥ 0 (points on or above the x-axis), and y ≤ -2x + 6 (points on or below the line y = -2x + 6). The first two constraints together restrict us to the first quadrant (where both x and y are non-negative). The third constraint further restricts us to points below the line y = -2x + 6, with a solid boundary since we have ≤. Choice C correctly identifies this as all points in the first quadrant that are on or below the line y = -2x + 6, including the boundary. Great work! Choice A incorrectly places the region in Quadrant II (where x < 0), Choice B describes the third quadrant (both x ≤ 0 and y ≤ 0), and Choice D has the wrong shading direction (above instead of below). For systems, think of it like finding what's allowed: each inequality restricts the plane, and the solution is where ALL the restrictions are met simultaneously—the overlapping shaded region. If you have y ≥ x and y ≤ -x + 4, the solution is the wedge-shaped region where both shadings overlap!

Question 4

Graph the inequality y>2x−1y > 2x - 1y>2x−1. Which description is correct for the boundary line and the shaded half-plane (solution region)?

  1. Solid boundary line y=2x−1y=2x-1y=2x−1; shade above the line
  2. Dashed boundary line y=−2x−1y=-2x-1y=−2x−1; shade above the line
  3. Dashed boundary line y=2x−1y=2x-1y=2x−1; shade above the line
  4. Solid boundary line y=2x−1y=2x-1y=2x−1; shade below the line
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. To graph a linear inequality like y > 2x + 1, we first graph the boundary line y = 2x + 1 (replacing the inequality with equals). Then we decide: is it a solid line (if the inequality includes 'or equal to,' like ≥ or ≤) or a dashed line (if it's strict, like > or <)? Finally, we shade the half-plane that makes the inequality true—above the line for y > or y ≥, below for y < or y ≤. For y > 2x - 1, start by graphing the boundary line y = 2x - 1 using points like (0, -1) and (1, 1); since it's a strict inequality (>), make the line dashed, then test a point like (0,0) in y > 2x - 1: 0 > -1 is true, so shade the side containing (0,0), which is above the line. Choice B correctly identifies the dashed boundary line y=2x-1 with shading above the line because the inequality is strict and requires y-values greater than the line. A common mistake is choosing a solid line like in choice A or D, but remember, strict inequalities use dashed lines to show boundary points aren't included—keep practicing to spot that difference! The solid-or-dashed rule is simple: if you see ≤ or ≥ (the inequality has a line underneath showing 'or equal to'), make the boundary line solid because those points are included. If you see < or > (strict inequality, no line underneath), make it dashed because boundary points don't count. Think: the line under the inequality symbol = solid line on the graph! For shading direction with y inequalities: y > [line] means 'y is greater than the line' = shade above (higher y-values). y < [line] means 'y is less than the line' = shade below (lower y-values). Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 5

Is point (2,1)(2,1)(2,1) in the solution region of the inequality y<x−2y < x - 2y<x−2?​​

  1. No, because 1≮01 \not< 01<0
  2. No, because 1≮−41 \not< -41<−4
  3. Yes, because 1<41 < 41<4
  4. Yes, because 1<01 < 01<0
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. The solution to a linear inequality is an entire region (a half-plane), not just a single point: every point in the shaded region makes the inequality true! This is different from linear equations, which have just one solution point where the lines cross. To check if point (2, 1) is in the solution region of y < x - 2, we substitute x = 2 and y = 1 into the inequality: 1 < 2 - 2, which simplifies to 1 < 0. Since 1 is NOT less than 0, this statement is false, so the point is not in the solution region. Choice C correctly identifies that the point is not in the solution region because 1 ≮ 0 (1 is not less than 0). Great work! Choices A and B incorrectly claim the point is in the solution (with A making the false claim that 1 < 0), while choice D uses the wrong calculation comparing 1 to -4 instead of 0. For shading direction with y inequalities: y > [line] means 'y is greater than the line' = shade above (higher y-values). y < [line] means 'y is less than the line' = shade below (lower y-values). Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 6

Graph the inequality 2x+y>42x + y > 42x+y>4. Which boundary line and shading are correct?​​

  1. Solid line 2x+y=42x+y=42x+y=4; shade the side that contains (0,0)(0,0)(0,0)
  2. Dashed line 2x+y=42x+y=42x+y=4; shade the side that contains (0,0)(0,0)(0,0)
  3. Dashed line 2x+y=42x+y=42x+y=4; shade the side that does not contain (0,0)(0,0)(0,0)
  4. Solid line 2x+y>42x+y>42x+y>4; shade above the line
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. To graph a linear inequality like 2x + y > 4, we first graph the boundary line 2x + y = 4 (replacing the inequality with equals). Then we decide: is it a solid line (if the inequality includes 'or equal to,' like ≥ or ≤) or a dashed line (if it's strict, like > or <)? Finally, we shade the half-plane that makes the inequality true—above the line for y > or y ≥, below for y < or y ≤. For 2x + y > 4, we have a strict inequality (>) so the boundary is dashed. To determine shading, test (0, 0): 2(0) + 0 > 4 gives 0 > 4, which is FALSE, so we shade the side that does NOT contain (0, 0). Choice C correctly identifies the dashed line 2x + y = 4 (because > is strict) and shading the side that does not contain (0, 0) (because the test point failed). Great work! Choices A and B incorrectly shade the side containing (0, 0) when our test shows it's not in the solution, and choice D nonsensically writes an inequality as the boundary line equation. Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 7

What is the boundary line for the inequality 3x−y≤63x - y \le 63x−y≤6?​​

  1. 3x−y≤63x - y \le 63x−y≤6
  2. 3x+y=63x + y = 63x+y=6
  3. 3x−y=63x - y = 63x−y=6
  4. 3x−y=−63x - y = -63x−y=−6
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. The boundary line for an inequality is the line you'd get if you changed the inequality to equals: for 3x - y ≤ 6, the boundary is 3x - y = 6. The line is dashed for strict inequalities (< or >) because points ON the line don't satisfy the inequality, and solid for ≤ or ≥ because boundary points ARE solutions. To find the boundary line for 3x - y ≤ 6, we simply replace the inequality symbol with an equals sign, giving us 3x - y = 6. Choice A correctly identifies 3x - y = 6 as the boundary line because this is exactly what we get when we change ≤ to =. Great work! Choice B incorrectly keeps the inequality symbol when we need just the equation, choice C has the wrong constant (-6 instead of 6), and choice D changes the minus to plus which alters the equation entirely. For shading direction with y inequalities: y > [line] means 'y is greater than the line' = shade above (higher y-values). y < [line] means 'y is less than the line' = shade below (lower y-values). Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 8

To determine shading for the inequality y>−3x+2y> -3x+2y>−3x+2, a student tests the point (0,0)(0,0)(0,0). What does this test point indicate?

  1. Since 0>−3(0)+20>-3(0)+20>−3(0)+2 is true, shade the half-plane that does not contain (0,0)(0,0)(0,0).
  2. Since 0>−3(0)+20>-3(0)+20>−3(0)+2 is false, shade the half-plane that does not contain (0,0)(0,0)(0,0).
  3. Since 0>−3(0)+20>-3(0)+20>−3(0)+2 is false, shade the half-plane that contains (0,0)(0,0)(0,0).
  4. Since 0>−3(0)+20>-3(0)+20>−3(0)+2 is true, shade the half-plane that contains (0,0)(0,0)(0,0).
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. To graph a linear inequality like y > -3x + 2, we first graph the boundary line y = -3x + 2 (replacing the inequality with equals). Then we decide: is it a solid line (if the inequality includes 'or equal to,' like ≥ or ≤) or a dashed line (if it's strict, like > or <)? Finally, we shade the half-plane that makes the inequality true—above the line for y > or y ≥, below for y < or y ≤. To use the test point method for y > -3x + 2, we substitute (0, 0) into the inequality: 0 > -3(0) + 2, which simplifies to 0 > 2. This statement is FALSE because 0 is not greater than 2. When the test point makes the inequality false, we shade the half-plane that does NOT contain the test point. Choice B correctly states that since 0 > -3(0) + 2 is false, we shade the half-plane that does not contain (0, 0). Great work! Choice A incorrectly evaluates 0 > 2 as true, Choice C has the right evaluation but wrong conclusion about shading, and Choice D has both parts backwards. Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 9

Which direction should be shaded for the inequality y<−3x+2y<-3x+2y<−3x+2?

  1. Shade above the line y=−3x+2y=-3x+2y=−3x+2.
  2. Shade below the line y=−3x+2y=-3x+2y=−3x+2.
  3. Shade to the right of the line y=−3x+2y=-3x+2y=−3x+2.
  4. Shade to the left of the line y=−3x+2y=-3x+2y=−3x+2.
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. To graph a linear inequality like y < -3x + 2, we first graph the boundary line y = -3x + 2 (replacing the inequality with equals). Then we decide: is it a solid line (if the inequality includes 'or equal to,' like ≥ or ≤) or a dashed line (if it's strict, like > or <)? Finally, we shade the half-plane that makes the inequality true—above the line for y > or y ≥, below for y < or y ≤. For y < -3x + 2, we need to shade where y-values are less than (smaller than) the expression -3x + 2. Since we want y-values that are smaller, we shade below the boundary line. Choice B correctly identifies shading below the line because y < -3x + 2 means we want all points where the y-coordinate is less than what the line gives us. Great work! Choice A incorrectly shades above (that would be for y > -3x + 2), and choices C and D use left/right language which doesn't apply to non-vertical lines. For shading direction with y inequalities: y > [line] means 'y is greater than the line' = shade above (higher y-values). y < [line] means 'y is less than the line' = shade below (lower y-values). Or use the test point method: pick (0, 0) if it's not on the line, substitute into the inequality, and if true, shade the side with (0, 0); if false, shade the other side!

Question 10

Graph the solution set to the system of inequalities {y>−2x+1y≤x+4\begin{cases}y> -2x+1\\y\le x+4\end{cases}{y>−2x+1y≤x+4​ Which description matches the intersection (overlap) of the half-planes?​

  1. The region above (but not including) y=−2x+1y=-2x+1y=−2x+1 and above (and including) y=x+4y=x+4y=x+4.
  2. The region below (but not including) y=−2x+1y=-2x+1y=−2x+1 and below (and including) y=x+4y=x+4y=x+4.
  3. The region above (but not including) y=−2x+1y=-2x+1y=−2x+1 and on or below y=x+4y=x+4y=x+4.
  4. The region on or below y=−2x+1y=-2x+1y=−2x+1 and above (but not including) y=x+4y=x+4y=x+4.
Explanation: This question tests your understanding of graphing linear inequalities and how the solution is represented as a shaded half-plane on the coordinate plane. A system of linear inequalities has a solution region that's the intersection (overlap) of all the individual half-planes: you graph each inequality, and where all the shaded regions overlap is where all the inequalities are satisfied at once. That intersection is your feasible region! For the system y > -2x + 1 and y ≤ x + 4, the first inequality has y greater than -2x + 1 (strict), so we shade above the dashed line y = -2x + 1. The second inequality has y less than or equal to x + 4, so we shade below (and on) the solid line y = x + 4. The solution region is where both conditions are met: above the first line AND on or below the second line. Choice C correctly describes this as the region above (but not including) y = -2x + 1 and on or below y = x + 4. Great work! Choices A and B incorrectly describe both regions as above or both as below, while choice D reverses the inequality types (making the first 'or equal to' and the second strict). For systems, think of it like finding what's allowed: each inequality restricts the plane, and the solution is where ALL the restrictions are met simultaneously—the overlapping shaded region. Pay careful attention to whether boundaries are included (solid) or excluded (dashed)!