Algebra Quiz: Graphing Linear Inequalities And Systems

What this quiz covers

This quiz focuses on Graphing Linear Inequalities And Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Yes, because $1 < 0$
2. Yes, because $1 < 4$
3. No, because $1 \not< 0$ (correct answer)
4. No, because $1 \not< -4$

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

1. The region below $y=2x$ and above $y=-x+6$.
2. The region above $y=2x$ and above $y=-x+6$.
3. The region above $y=2x$ and on or below $y=-x+6$. (correct answer)
4. The region on or below $y=2x$ and on or below $y=-x+6$.

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

1. All points in the first quadrant that are on or below the line $y=-2x+6$, including the boundary. (correct answer)
2. All points with $x\le 0$ and $y\le 0$ that are below the line $y=-2x+6$.
3. All points in Quadrant II on or above the line $y=-2x+6$.
4. All points in the first quadrant that are on or above the line $y=-2x+6$, 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

1. Solid boundary line $y=2x-1$; shade above the line
2. Dashed boundary line $y=-2x-1$; shade above the line
3. Dashed boundary line $y=2x-1$; shade above the line (correct answer)
4. Solid boundary line $y=2x-1$; 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

1. No, because $1 \not< 0$ (correct answer)
2. No, because $1 \not< -4$
3. Yes, because $1 < 4$
4. Yes, because $1 < 0$

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

1. Solid line $2x+y=4$; shade the side that contains $(0,0)$
2. Dashed line $2x+y=4$; shade the side that contains $(0,0)$
3. Dashed line $2x+y=4$; shade the side that does not contain $(0,0)$ (correct answer)
4. Solid line $2x+y>4$; 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

1. $3x - y \le 6$
2. $3x + y = 6$
3. $3x - y = 6$ (correct answer)
4. $3x - y = -6$

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

1. Since $0>-3(0)+2$ is true, shade the half-plane that does not contain $(0,0)$.
2. Since $0>-3(0)+2$ is false, shade the half-plane that does not contain $(0,0)$. (correct answer)
3. Since $0>-3(0)+2$ is false, shade the half-plane that contains $(0,0)$.
4. Since $0>-3(0)+2$ is true, shade the half-plane that contains $(0,0)$.

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

1. Shade above the line $y=-3x+2$.
2. Shade below the line $y=-3x+2$. (correct answer)
3. Shade to the right of the line $y=-3x+2$.
4. Shade to the left of the line $y=-3x+2$.

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

1. The region above (but not including) $y=-2x+1$ and above (and including) $y=x+4$.
2. The region below (but not including) $y=-2x+1$ and below (and including) $y=x+4$.
3. The region above (but not including) $y=-2x+1$ and on or below $y=x+4$. (correct answer)
4. The region on or below $y=-2x+1$ and above (but not including) $y=x+4$.

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)!

Question 11

1. All points left of the solid vertical line $x=-1$ and above the solid horizontal line $y=2$
2. All points right of the solid vertical line $x=-1$ and below the solid horizontal line $y=2$ (correct answer)
3. All points right of the dashed vertical line $x=-1$ and below the dashed horizontal line $y=2$
4. All points left of the dashed vertical line $x=-1$ and above the dashed horizontal line $y=2$

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. 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 $x \ge -1$, we have a vertical boundary line $x = -1$ that's solid (because of $\ge$), and we shade to the right where x-values are greater. For $y \le 2$, we have a horizontal boundary line $y = 2$ that's solid (because of $\le$), and we shade below where y-values are less. The solution is the intersection: right of $x = -1$ AND below $y = 2$. Choice B correctly identifies all points right of the solid vertical line $x = -1$ and below the solid horizontal line $y = 2$, with both boundaries solid because both inequalities include equality. Great work! Choice A has the wrong directions (left and above), choice C incorrectly uses dashed lines when $\ge$ and $\le$ require solid lines, and choice D has wrong directions and wrong line types. 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 $y \ge x$ and $y \le -x + 4$, the solution is the wedge-shaped region where both shadings overlap!

Question 12

1. $2x+y>6$
2. $2x+y=6$ (correct answer)
3. $2x+y\le 6$
4. $2x-y=6$

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 2x + y ≥ 6, the boundary is 2x + 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 2x + y ≥ 6, we simply replace the inequality symbol (≥) with an equals sign (=). This gives us the equation 2x + y = 6, which represents the line that separates the solution region from the non-solution region. Choice B correctly identifies 2x + y = 6 as the boundary line because this is the equation we get when we change the inequality to an equation. Great work! Choices A and C keep the inequality symbols (which don't belong in a boundary line equation), and choice D incorrectly changes the plus to minus. The boundary line for an inequality is always found by replacing the inequality symbol with equals—it's that simple! This line will be drawn as solid (for ≤ or ≥) or dashed (for < or >) depending on whether the boundary points are included in the solution.

Question 13

1. $2x+y<6$
2. $2x+y=6$ (correct answer)
3. $2x+y\le 6$
4. $2x-y=6$

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 2x + y < 6, the boundary is 2x + 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 2x + y < 6, we simply replace the inequality symbol with an equals sign, giving us 2x + y = 6. This is the line that separates the plane into two half-planes, one of which will be our solution region. Choice B correctly identifies 2x + y = 6 as the boundary line. Great work! Choice A gives the original inequality, not the boundary line; choice C changes the inequality type but still isn't just the boundary; and choice D incorrectly changes the plus to minus. The boundary line for an inequality is always found by replacing the inequality symbol with equals—it's that simple! Whether the line is drawn solid or dashed depends on the original inequality symbol, but the equation of the boundary line itself is always the equality version.

Question 14

1. $3x - y = 6$ (correct answer)
2. $3x - y \le 6$
3. $3x - y = -6$
4. $3x + y = 6$

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 15

1. $y = -2x - 5$
2. $y = 2x + 5$
3. $y > -2x + 5$
4. $y = -2x + 5$ (correct answer)

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 $2x + y > 6$, the boundary is $2x + y = 6$. 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. For $2x + y > 5$, the boundary is $2x + y = 5$; to write in slope-intercept form, solve for y: $y = -2x + 5$ (subtract $2x$ from both sides). Choice B correctly identifies $y = -2x + 5$ because solving $2x + y = 5$ for y gives that equation, with slope $-2$ and y-intercept $5$. Great work! Choice A has a positive slope, but it should be negative since it's $+2x$ moving to $-2x$; choice C adds a negative intercept incorrectly; and D keeps the inequality, but the boundary is the equality version—nice try, but remembering to set to equals and solve for y will fix that. 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!

Question 16

1. $(0,0)$ satisfies the inequality, so shade the side containing the origin
2. $(0,0)$ satisfies the inequality, so shade the side not containing the origin
3. $(0,0)$ does not satisfy the inequality, so shade the side not containing the origin (correct answer)
4. $(0,0)$ does not satisfy the inequality, so shade the side containing the origin

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 > x + 3, graph dashed boundary y = x + 3; testing (0,0): 0 > 0 + 3 is false, so shade the opposite side, not containing (0,0). Choice C correctly identifies that (0,0) does not satisfy, so shade the side not containing the origin. If you selected A or B assuming it satisfies, double-check the substitution—0 > 3 is false, so it's a great reminder to always verify with numbers. 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 17

1. Yes, because $3\ge 2(-1)$ and $3\le (-1)+4$ are both true. (correct answer)
2. Yes, because at least one of the inequalities is true.
3. No, because $3\ge 2(-1)$ is false.
4. No, because $3\le (-1)+4$ is false.

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 (-1, 3) is in the solution set of the system, we need to verify BOTH inequalities. For y ≥ 2x: substitute to get 3 ≥ 2(-1), which gives 3 ≥ -2. This is TRUE. For y ≤ x + 4: substitute to get 3 ≤ (-1) + 4, which gives 3 ≤ 3. This is also TRUE (remember ≤ includes equals). Since both inequalities are satisfied, the point is in the solution set. Choice A correctly states that (-1, 3) is in the solution set because both 3 ≥ 2(-1) and 3 ≤ (-1) + 4 are true. Great work! Choice B is wrong because BOTH inequalities must be true for a system, not just one. Choices C and D incorrectly evaluate the inequalities—both statements are actually true when calculated correctly. 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 18

1. $x-2y=8$ (correct answer)
2. $x-2y\le 8$
3. $x+2y=8$
4. $x-2y<8$

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 x - 2y < 8, the boundary is x - 2y = 8. 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 x - 2y < 8, we replace the inequality symbol (<) with an equals sign (=). This gives us x - 2y = 8, which is the line that separates the solution region from the non-solution region. Choice A correctly identifies x - 2y = 8 as the boundary line because this is what we get when we change the inequality to an equation. Great work! Choices B and D keep inequality symbols (which don't belong in a boundary line equation), and choice C incorrectly changes the minus to plus. The boundary line for an inequality is always found by replacing the inequality symbol with equals—it's that simple! Remember, the boundary line is an equation (with =), not an inequality, regardless of whether it will be drawn solid or dashed.

Question 19

1. Points with $x\le 10$, $y\le 15$, and $x+y\ge 80$
2. Points with $x\ge 10$, $y\ge 15$, and $x+y\le 80$ (intersection of all three half-planes) (correct answer)
3. Points with $x\ge 10$ or $y\ge 15$ or $x+y\le 80$ (union of the half-planes)
4. Points with $x\le 10$, $y\ge 15$, and $x+y\le 80$

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 x + y ≤ 80 (solid line x+y=80, shade below), x ≥ 10 (solid vertical, shade right), y ≥ 15 (solid horizontal, shade above), the feasible region is the polygon where all overlap: x≥10, y≥15, x+y≤80. Choice B correctly identifies points with x≥10, y≥15, and x+y≤80 as the intersection of all three half-planes. Choice C says 'or' which is union, but systems require all conditions (intersection)—great effort, but remember 'and' for systems. 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 20

1. All points in Quadrant I on or above the line $x+y=6$
2. All points in Quadrant I above the line $x+y=6$
3. All points on or below the line $x+y=6$ in all four quadrants
4. All points in Quadrant I on or below the line $x+y=6$ (correct answer)

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 x ≥ 0 (solid vertical line at x=0, shade right), y ≥ 0 (solid horizontal at y=0, shade above), and x + y ≤ 6 (solid line x + y = 6, shade below since solving for y gives y ≤ -x + 6), the overlap is in Quadrant I, on or below the line x + y = 6. Choice B correctly identifies all points in Quadrant I on or below the line x+y=6 because that's the intersection of the three half-planes in the first quadrant. Great work! Choice C includes all quadrants below the line, but x ≥ 0 and y ≥ 0 restrict it to Quadrant I—adding those non-negativity constraints is key; don't fret, visualizing each shade helps. 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!