Graphing Linear Inequalities and Systems - Algebra 1

Is point $(2,1)$ in the solution region of the inequality $y> x-1$ ?
Answer: No, because $1>2-1$ is false. 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 - 1, we substitute x = 2 and y = 1 into the inequality. This gives us: 1 > 2 - 1, which simplifies to 1 > 1. Since 1 is not greater than 1 (they're equal), this statement is false. Choice C correctly states that (2, 1) is not in the solution region because 1 > 2 - 1 is false. Great work! Choice A incorrectly evaluates the arithmetic (2 - 1 = 1, not something less than 1), Choice B uses ≥ instead of the given >, and Choice D introduces ≤ which isn't part of the original inequality. 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!
Graph the inequality $y > 2x - 1$ . Should the boundary line be solid or dashed, and which half-plane should be shaded?
Answer: Dashed boundary line $y=2x-1$ ; shade above the line 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, which has a slope of 2 and y-intercept of -1; since it's a strict inequality (>), use a dashed line, and because it's y >, shade above the line where y-values are greater. Choice B correctly identifies dashed boundary line y=2x-1; shade above the line because the inequality is strict and requires shading the region with higher y-values. Great work! A common mistake is choosing a solid line, like in choices A and D, but remember, solid lines are only for inequalities that include equality (≥ or ≤), so that's why those are incorrect—keep practicing to avoid that mix-up. 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!
A system of three linear inequalities creates a triangular solution region. If one vertex of this triangle is at $(2, 5)$ and the three boundary lines are $y = x + 3$ , $y = -2x + 9$ , and $x = 0$ , what type of boundary does the line $x = 0$ have in this system?
Answer: Solid boundary with inequality $x \geq 0$ to create the triangular region First, verify (2,5) is at intersection of two lines: y = x + 3 gives 5 = 2 + 3 ✓. y = -2x + 9 gives 5 = -4 + 9 ✓. So (2,5) is where these lines intersect. For a triangular region in the first quadrant with vertex at (2,5), the third boundary x = 0 must use x ≥ 0 to include the region to the right of the y-axis. The boundary should be solid since we need the triangle to have definite vertices. Choice B: Dashed boundary would exclude the y-axis. Choice C: x ≤ 0 would put triangle in second quadrant. Choice D: We have sufficient information to determine this.
To determine shading for the inequality $y> -3x+2$ , a student tests the point $(0,0)$ . What does this test point indicate?
Answer: Since $0>-3(0)+2$ is false, shade the half-plane that does not contain $(0,0)$ . 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!
Graph the solution set to the system $\begin{cases} y \ge x + 1 \\ y < -x + 5 \end{cases}$ Which description best matches the solution region?
Answer: The region above (or on) $y=x+1$ and below (but not on) $y=-x+5$ 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 $y \ge x + 1$ (solid line, shade above or on) and $y < -x + 5$ (dashed line, shade below), the overlap is above or on the first line and below but not on the second. Choice C correctly identifies the region above (or on) $y=x+1$ and below (but not on) $y=-x+5$ because it matches the inclusive $\ge$ for the first and strict $<$ for the second. If you picked B, that reverses the directions—gently remember to check shading for each inequality separately before finding the overlap. 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!
Which direction should be shaded for the inequality $y<-3x+2$ ?
Answer: Shade below the line $y=-3x+2$ . 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!
Is point $(2,3)$ in the solution region of the inequality $y< x+1$ ?
Answer: No, because $3<2+1$ is false. 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, 3) is in the solution region of y < x + 1, we substitute x = 2 and y = 3 into the inequality: 3 < 2 + 1, which simplifies to 3 < 3. This statement is false because 3 is not less than 3 (it's equal to 3). Therefore, the point (2, 3) is not in the solution region. Choice C correctly identifies that the point is not in the solution region because 3 < 2 + 1 is false. Great work! Choice A incorrectly claims 3 < 3 is true, choice B uses the wrong inequality symbol (≤), and choice D makes an incorrect general statement about boundary 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!
Is point $(0,0)$ in the solution region of the system $\begin{cases} y>2x-3 \\ y\le x+1 \end{cases}$ ?
Answer: Yes, because it satisfies both inequalities. 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! For a system, a point must satisfy ALL inequalities to be in the solution region. To check if (0, 0) is in the solution region, we test both inequalities. For y > 2x - 3: substituting gives 0 > 2(0) - 3, which is 0 > -3, TRUE. For y ≤ x + 1: substituting gives 0 ≤ 0 + 1, which is 0 ≤ 1, TRUE. Since both inequalities are satisfied, the point is in the solution region. Choice A correctly identifies that (0, 0) is in the solution region because it satisfies both inequalities: 0 > -3 and 0 ≤ 1 are both true statements. Great work! Choices B, C, and D incorrectly claim the point fails one or both inequalities when our calculations show both are satisfied. 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! For systems, the point must satisfy ALL inequalities—if even one fails, the point is outside the solution region.
What is the boundary line for the inequality $2x + y > 5$ (written in slope-intercept form)?
Answer: $y = -2x + 5$ 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!
Graph the solution set to the system $\begin{cases} x \ge 0 \\ y \ge 0 \\ y \le -x + 4 \end{cases}$ Which description is correct?
Answer: All points in the first quadrant that are on or below the line $y = -x + 4$ 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 $x \ge 0$ (solid vertical at x=0, shade right), $y \ge 0$ (solid horizontal at y=0, shade above), $y \le -x + 4$ (solid line, shade below or on), the overlap is the first quadrant triangle on or below $y = -x + 4$ . Choice B correctly identifies all points in the first quadrant that are on or below the line $y = -x + 4$ because it captures the intersection in the non-negative region. Choice A mentions Quadrant II, but that's x<0—gently confirm quadrants: first is $x \ge 0$ , $y \ge 0$ . 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!
Graph the inequality $2x + y > 4$ . Which boundary line and shading are correct?
Answer: Dashed line $2x+y=4$ ; shade the side that does not contain $(0,0)$ 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!
What is the boundary line for the inequality $2x+y\ge 6$ ?
Answer: $2x+y=6$ 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.
Describe the solution region for the system $\begin{cases} y \ge 0 \\ y \le -x + 3 \end{cases}$
Answer: All points above (or on) the $x$ -axis and below (or on) $y=-x+3$ 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 $y \ge 0$ , this means y-values must be non-negative, which is all points on or above the x-axis (where $y = 0$ ). For $y \le -x + 3$ , we shade on or below the line $y = -x + 3$ . The solution region is where BOTH conditions are met: above (or on) the x-axis AND below (or on) the line $y = -x + 3$ . Choice C correctly identifies all points above (or on) the $x$ -axis and below (or on) $y = -x + 3$ , properly including the boundaries since both inequalities use $\ge$ and $\le$ . Great work! Choice A excludes the boundaries incorrectly, choice B has the wrong region for the second inequality (above instead of below), and choice D has the wrong region for the first inequality (below x-axis instead of above). 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!
What is the boundary line for the inequality $2x-y>3$ ?
Answer: $2x-y=3$ 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 > 3, the boundary is 2x - y = 3. 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 > 3, we simply replace the inequality symbol (>) with an equals sign (=). This gives us the equation 2x - y = 3, which represents the line that separates the plane into two half-planes. Choice B correctly identifies 2x - y = 3 as the boundary line. Great work! Choices A, C, and D incorrectly keep inequality symbols—remember, a boundary line is always an equation (with =), not an inequality. The boundary line itself will be dashed when graphed because we have a strict inequality (>). 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!
What is the boundary line for the inequality $3x - y \le 6$ ?
Answer: $3x - y = 6$ 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!
Is point $(2,1)$ in the solution region of the inequality $y < x - 2$ ?
Answer: No, because $1 \not< 0$ 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!
What is the boundary line for the inequality $2x+y<6$ ?
Answer: $2x+y=6$ 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.
Is point $(2,1)$ in the solution region of the inequality $y\ge x-1$ ?
Answer: Yes, because $1\ge 2-1$ is true. 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 - 1, we substitute x = 2 and y = 1 into the inequality: 1 ≥ 2 - 1, which simplifies to 1 ≥ 1. Since this is true (1 equals 1, and ≥ includes equality), the point is in the solution region. Choice A correctly identifies that (2, 1) is in the solution region because 1 ≥ 2 - 1 evaluates to 1 ≥ 1, which is true. Great work! Choice B incorrectly claims this is false, choice C gives a wrong geometric reason (the point is actually on the line, not below it), and choice D is irrelevant since the boundary would be solid for ≥. To test if a point is in the solution region, simply substitute its coordinates into the inequality. If the resulting statement is true, the point is in the solution region. If false, it's not. This works for any point and any linear inequality!
What is the boundary line for the inequality $x-2y<8$ ?
Answer: $x-2y=8$ 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.
Graph the solution set to the system of inequalities $\begin{cases} y \geq x + 1 \\ y < -2x + 6 \end{cases}$ Which description matches the solution region?
Answer: The region above (or on) $y=x+1$ and below (but not on) $y=-2x+6$ 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 $y \geq x + 1$ and $y < -2x + 6$ , graph $y = x + 1$ as solid (because $\geq$ ) and shade above ( $y \geq$ ), then graph $y = -2x + 6$ as dashed (because $<$ strict) and shade below ( $y <$ ); the solution is where these overlap—above or on $y = x + 1$ and below (but not on) $y = -2x + 6$ . Choice C correctly identifies the region above (or on) $y=x+1$ and below (but not on) $y=-2x+6$ because it matches the shading and boundary types for each inequality. Great work! Choice B says below $y=x+1$ , but that's the opposite of $y \geq$ which shades above—easy to mix up, but remembering $y >$ means above helps; keep practicing systems to get comfortable with overlaps. 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 $y \geq x$ and $y \leq -x + 4$ , the solution is the wedge-shaped region where both shadings overlap!

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