Free math worksheets

- Test the ordered pair in each inequality separately. - A system requires every condition to be true at the same time. - Equality identifies a boundary point; a strict comparison places the point off the boundary.

1. For inequality (I), \(2+1=3<4\), so the point satisfies the inequality strictly. 2. For inequality (II), \(1>2-2=0\), so the point satisfies the inequality strictly. 3. Since the point satisfies both inequalities, it is a solution of the system.

Yes. The point \((2, 1)\) satisfies both inequalities strictly, so it is a solution of the system.

- Read each boundary line separately. - Use line style to decide whether equality is included. - The system is represented by the overlap of the two shaded conditions.

1. The vertical boundary is \(x=-2\). It is dashed, and the shading is to the right, so the first inequality is \(x>-2\). 2. The horizontal boundary is \(y=3\). It is solid, and the shading is below, so the second inequality is \(y\le3\). 3. Any point in the overlap satisfies both inequalities; for example, \((0, 0)\) is a solution.

The system is \(x>-2\) and \(y\le3\). One solution is \((0, 0)\).

- Combine the two opposite inclusive bounds on x. - Keep the remaining y-condition after the x-coordinate is fixed. - Describe the resulting one-dimensional set geometrically.

1. The first two inequalities require \(x\) to be both at least and at most \(2\), so \(x=2\). 2. The third inequality allows every y-value with \(y\ge-1\). 3. Therefore, the solution set is a vertical ray beginning at \((2, -1)\) and extending upward.

The solution set is \(\{(2, y)\mid y\ge-1\}\), a vertical ray with endpoint \((2, -1)\) included.

- A solution must lie in the overlap of all shaded conditions. - Included boundary segments belong to the solution set. - Check points near the slanted boundary against its limiting equation.

1. Points \(A=(0, 6)\), \(B=(1, 4)\), and \(C=(3, 0)\) lie on the included boundary of the shaded intersection, so they are solutions. 2. Point \(D=(-1, 2)\) lies outside the shaded intersection because its x-coordinate is negative. 3. Point \(E=(2, 3)\) lies above the slanted boundary because \(2\cdot2+3=7>6\), so it is not a solution.

Points \(A\), \(B\), and \(C\) are solutions: \((0, 6)\), \((1, 4)\), and \((3, 0)\).

- Rewrite the two inequalities so their left sides can be compared directly. - Determine which condition is stricter. - A condition is redundant when removing it does not enlarge the common solution set.

1. Divide inequality (II) by \(2\) to get \(x+y\le7\). 2. Every point satisfying inequality (I), \(x+y\le6\), also satisfies \(x+y\le7\). 3. Therefore, inequality (II) is redundant, and the system represents the half-plane \(x+y\le6\).

The inequality \(2x+2y\le14\) is redundant. The solution set is \(x+y\le6\).

- Read the horizontal and vertical side lengths from the coordinate bounds. - Use the rectangle formulas for area and perimeter. - A set is bounded when it cannot extend indefinitely in any direction.

1. The x-values run from \(0\) to \(4\), so the rectangle has width \(4\). 2. The y-values run from \(0\) to \(3\), so the rectangle has height \(3\). 3. Its area is \(4\cdot3=12\) square units, and its perimeter is \(2(4+3)=14\) units. 4. Both coordinates have lower and upper bounds, so the solution set is bounded.

The area is \(12\) square units, the perimeter is \(14\) units, and the solution set is bounded.

- Combine the information supplied by the two coordinate restrictions. - Compare the resulting statement about the sum with the third condition. - Look for a direct contradiction involving the same expression.

1. From \(x\ge0\) and \(y\ge0\), adding the inequalities gives \(x+y\ge0\). 2. The third inequality requires \(x+y<0\). 3. The same sum cannot be both at least \(0\) and less than \(0\), so the system is inconsistent.

The system has no solutions.

- Simplify the scaled inequality before comparing the two conditions. - Ask when one expression can be both at least and at most the same value. - Describe the resulting equality geometrically.

1. Divide inequality (II) by \(2\) to get \(x+y\le4\). 2. The system therefore requires \(x+y\ge4\) and \(x+y\le4\). 3. Both conditions hold exactly when \(x+y=4\). 4. The solution set is the entire common boundary line.

The solution set is every point on the line \(x+y=4\).

- Interpret each comparison relative to its own boundary line. - Separate the shading direction from the boundary style. - The common set must satisfy a lower and an upper y-bound simultaneously.

1. The first inequality selects points on or above \(y=2x-1\), so that boundary is solid. 2. The second inequality selects points below \(y=2x+3\), not above it, so that boundary is dashed. 3. The correct overlap is the strip \(2x-1\le y<2x+3\) between the parallel lines.

Priya shaded the wrong side of \(y=2x+3\). The correct solution set is \(2x-1\le y<2x+3\), with the lower boundary solid and the upper boundary dashed.

- Simplify the coordinate sum before treating the other restrictions. - Convert each coordinate condition into an interval for the parameter. - Intersect the intervals, keeping included endpoints.

1. For every \(t\), the point satisfies \(x+y=t+(4-t)=4\), so the first inequality holds with equality. 2. The condition \(x\ge1\) gives \(t\ge1\). 3. The condition \(y\ge0\) gives \(4-t\ge0\), so \(t\le4\). 4. The common range is \(1\le t\le4\).

\(1\le t\le4\)

- Treat the repeated expression as one quantity. - Compare its required lower and upper bounds. - Check whether the interval between those bounds actually exists.

1. The first inequality requires \(x+y\) to be at most \(4\). 2. The second requires the same expression to be at least \(6\). 3. No real value can satisfy both restrictions because \(6>4\). 4. The half-planes face away from each other, so there is no overlap.

The system has no solutions. A strip would require the lower bound to be no greater than the upper bound.

- Compare the slopes and vertical separation of the two boundaries. - Look for an x-coordinate that makes the parameter term disappear. - Choose a y-value between the two resulting bounds.

1. The boundary lines have the same slope \(c\) and y-intercepts \(1\) and \(5\), so the lower boundary is always \(4\) units below the upper boundary. 2. For every x-value, all y-values satisfying \(cx+1\le y\le cx+5\) are solutions, so the system has infinitely many solutions. 3. At \(x=0\), the conditions become \(y\ge1\) and \(y\le5\). 4. Choosing \(y=2\) shows that \((0, 2)\) is a solution for every real \(c\).

For every real \(c\), the solution set is the strip between the two parallel lines, including both boundaries, so it contains infinitely many points. The point \((0, 2)\) is a solution for every value of \(c\).

- Treat the two inequalities separately before finding their overlap. - Equality versus strict inequality determines whether each boundary is included. - Test a simple point on either side of a boundary if you are unsure which side satisfies its inequality.

1. The inequality \(x\ge1\) has the vertical boundary \(x=1\). Equality is included, so the boundary is solid, and the satisfying side is to the right. 2. The inequality \(y<3\) has the horizontal boundary \(y=3\). It is strict, so the boundary is dashed, and the satisfying side is below. 3. The overlap is therefore to the right of \(x=1\) and below \(y=3\), which is panel a). 4. For example, \((2,2)\) lies in that overlap.

Panel a). The line \(x=1\) is solid with shading to its right; the line \(y=3\) is dashed with shading below it. One feasible point is \((2,2)\).

- Vertices occur where two boundary lines meet. - Read the axis intercepts of the slanted boundary. - Solid boundaries belong to the solution set.

1. The horizontal and vertical boundary segments meet at \((0, 0)\). 2. The slanted boundary meets the x-axis at \((8, 0)\) and the y-axis at \((0, 8)\). 3. All three boundaries are solid, so all edges and vertices are included.

The vertices are \((0, 0)\), \((8, 0)\), and \((0, 8)\). Every edge is included.

- Find the x-value where the two boundary outputs are equal. - On each side of that x-value, compare which line has the smaller y-value. - Because both inequalities are upper bounds, the lower line controls the overlap.

1. Set the boundary expressions equal: \(x+4=-2x+1\). 2. Solving gives \(3x=-3\), so \(x=-1\) and \(y=3\). The boundaries intersect at \((-1, 3)\). 3. For \(x<-1\), \(x+4<-2x+1\), so inequality (I) provides the tighter upper bound. 4. For \(x>-1\), \(-2x+1<x+4\), so inequality (II) provides the tighter upper bound.

The boundaries intersect at \((-1, 3)\). For \(x<-1\), \(y\le x+4\) is tighter. For \(x>-1\), \(y\le-2x+1\) is tighter.

- Translate the duration limit and the segment-count requirement separately. - Include restrictions that make sense for quantities being counted. - Test the proposed plan in every condition, not just the time limit.

1. The time limit gives \(4x+7y\le56\). 2. The segment requirement gives \(x+y\ge8\), with \(x\ge0\), \(y\ge0\), and whole-number counts. 3. For \((5, 4)\), the time is \(4\cdot5+7\cdot4=48\le56\). 4. The total number of segments is \(5+4=9\ge8\), so the plan is feasible.

The system is \(4x+7y\le56\), \(x+y\ge8\), \(x\ge0\), and \(y\ge0\), with whole-number values. The plan \((5, 4)\) is feasible.

- Analyze each system independently rather than relying on the other matches. - Identify boundary equations first, then compare the shaded side of each boundary. - A system's graph is the overlap of all of its inequalities.

1. System A requires the first quadrant and points at or below \(x+y=4\). Graph 2 has the two coordinate-axis boundaries and the descending boundary \(x+y=4\), with their common triangular region shaded. 2. System B requires points between the parallel lines \(y=x-1\) and \(y=x+2\). Graph 3 shows exactly that strip. 3. System C requires points left of the vertical line \(x=2\) and above the horizontal line \(y=-1\). Graph 1 shows that overlap.

A) Graph 2 — first-quadrant region below \(x+y=4\). B) Graph 3 — strip between \(y=x-1\) and \(y=x+2\). C) Graph 1 — region left of \(x=2\) and above \(y=-1\).

- Use one coordinate restriction to bound the expression in the third inequality. - Compare that bound with the separate restriction on y. - Decide whether the third condition removes any point already allowed by the first two.

1. From \(x\le-1\), it follows that \(x+1\le0\). 2. From \(y\ge0\), every solution has \(y\ge0\ge x+1\). 3. Therefore, every point satisfying the first two inequalities automatically satisfies \(y\ge x+1\). 4. The inequality \(y\ge x+1\) is redundant.

\(y\ge x+1\) is redundant.

- Find where the two boundary lines meet. - Compare the lower and upper bounds, preserving the strict condition. - Check separately whether the intersection x-value itself permits any y-value.

1. Set the boundary expressions equal: \(x+2=-x+6\). 2. Solving gives \(2x=4\), so the boundaries meet at \((2, 4)\). 3. A solution requires \(x+2<-x+6\) because the lower bound is strict. 4. This simplifies to \(x<2\). At \(x=2\), no y-value can be both greater than and at most \(4\).

The boundaries intersect at \((2, 4)\), and solutions exist only for \(x<2\).

- On each axis, compare the two upper bounds to see which one becomes active first. - Solve the two slanted boundary equations together for the non-axis vertex. - At a vertex, identify which boundary equations hold with equality.

1. The coordinate axes meet at \((0, 0)\). 2. On the x-axis, \(x+y=6\) gives \((6, 0)\), while \(x+2y=8\) would allow \((8,0)\). Thus \((6,0)\) is the x-axis vertex, with active boundaries \(y=0\) and \(x+y=6\). 3. The two slanted boundaries intersect where \(x+y=6\) and \(x+2y=8\). Subtracting gives \(y=2\), then \(x=4\), so \((4,2)\) is a vertex with both slanted boundaries active. 4. On the y-axis, \(x+2y=8\) gives \((0,4)\), while \(x+y=6\) would allow \((0,6)\). Thus \((0,4)\) is the y-axis vertex, with active boundaries \(x=0\) and \(x+2y=8\).

The vertices are \((0, 0)\), \((6, 0)\), \((4, 2)\), and \((0, 4)\). At \((6,0)\): \(y=0\) and \(x+y=6\). At \((4,2)\): \(x+y=6\) and \(x+2y=8\). At \((0,4)\): \(x=0\) and \(x+2y=8\).

- Match each edge to the inequality that creates it. - An inclusive edge can still lose endpoints that violate other strict conditions. - Test each corner in all three inequalities.

1. The vertical boundary \(x=1\) and horizontal boundary \(y=2\) are excluded because their inequalities are strict. 2. The diagonal boundary \(x+y=7\) is included only where \(x>1\) and \(y>2\). 3. The point \((1, 2)\) fails both strict coordinate inequalities. 4. The point \((1, 6)\) fails \(x>1\), and \((5, 2)\) fails \(y>2\). 5. Therefore, none of the three corner points is included, although the open portion of the diagonal between the latter two points is included.

The boundaries \(x=1\) and \(y=2\) are excluded. The diagonal \(x+y=7\) is included for points with \(1<x<5\). None of the three corner points is a solution.

- Find intersections of boundary lines that satisfy every inequality. - A boundary is active at a vertex when its inequality holds with equality there. - Check the horizontal and slanted upper bounds separately.

1. The coordinate axes meet at \((0, 0)\). 2. On the x-axis, \(x+y=6\) gives the vertex \((6, 0)\). 3. The boundaries \(y=4\) and \(x+y=6\) intersect at \((2, 4)\). 4. The boundary \(y=4\) meets the y-axis at \((0, 4)\). 5. Thus \(y=4\) is active at \((0,4)\) and \((2,4)\), while \(x+y=6\) is active at \((6,0)\) and \((2,4)\).

The vertices are \((0, 0)\), \((6, 0)\), \((2, 4)\), and \((0, 4)\). The boundary \(y=4\) is active at \((0,4)\) and \((2,4)\). The boundary \(x+y=6\) is active at \((6,0)\) and \((2,4)\).

- Test each point against all three inequalities. - A vertex lies where at least two active boundary lines meet. - Equality shows which boundary lines contain a solution point.

1. \((1, 1)\) lies on both \(x=1\) and \(y=1\), so it is a vertex. 2. \((2, 2)\) satisfies all three inequalities strictly, so it is a solution on no boundary. 3. \((3, 1)\) satisfies the system and lies only on \(y=1\), so it is a solution on exactly one boundary. 4. \((0, 3)\) fails \(x\ge1\), so it is not a solution.

\((1, 1)\): vertex \((2, 2)\): solution on no boundary \((3, 1)\): solution on exactly one boundary \((0, 3)\): not a solution

- Read each boundary equation from its orientation and intercepts. - Use the shaded side to choose each comparison direction. - A set is bounded when the boundaries enclose it in every direction.

1. The vertical boundary is \(x=1\), and the shading is to its right, so \(x\ge1\). 2. The horizontal boundary is \(y=0\), and the shading is above it, so \(y\ge0\). 3. The slanted boundary passes through \((1, 4)\) and \((5, 0)\), so it is \(x+y=5\). The shading is below it, so \(x+y\le5\). 4. The three included boundaries enclose a finite triangle, so the solution set is bounded.

The system is \(x\ge1\), \(y\ge0\), and \(x+y\le5\). The solution set is bounded.

- Find each axis intercept by setting the other coordinate to zero. - Test the origin to decide which side of the slanted boundary is selected. - Check whether the constraints impose any upper limit in either coordinate direction.

1. The boundary \(x+2y=6\) meets the x-axis at \((6, 0)\) and the y-axis at \((0, 3)\). 2. The inequality \(x+2y\ge6\) selects the side away from the origin. 3. Together with nonnegative coordinates, the solution set lies in the first quadrant on or beyond the slanted line. 4. It extends indefinitely to the right and upward, so it is unbounded.

The slanted boundary intercepts are \((6, 0)\) and \((0, 3)\). The solution set is the first-quadrant region on or beyond that line, and it is unbounded.

- Find where each slanted boundary meets the x-axis. - Solve the two slanted boundary equations together. - Match each pair of adjacent vertices to the boundary equation they share.

1. The boundary \(y=2x+4\) meets \(y=0\) at \((-2, 0)\). 2. The boundary \(y=-x+7\) meets \(y=0\) at \((7, 0)\). 3. The two slanted boundaries intersect where \(2x+4=-x+7\). This gives \(x=1\) and \(y=6\), so the third vertex is \((1,6)\). 4. The three sides lie on \(y=0\), \(y=2x+4\), and \(y=-x+7\), respectively.

The vertices are \((-2, 0)\), \((7, 0)\), and \((1, 6)\). The sides lie on \(y=0\), \(y=2x+4\), and \(y=-x+7\).

- Find all valid intersections of the boundary lines. - A constraint is active along an edge when its boundary equation holds throughout that edge. - Match each slanted pair of vertices to the boundary equation they share.

1. On the x-axis, \(x+y=5\) gives \((5,0)\), and \(x+2y=8\) gives \((8,0)\). 2. The two slanted boundaries intersect where \(x+y=5\) and \(x+2y=8\). Subtracting gives \(y=3\), then \(x=2\), so the third vertex is \((2,3)\). 3. No point on the y-axis satisfies both slanted inequalities, so the vertices are \((5,0)\), \((8,0)\), and \((2,3)\). 4. The edge from \((5,0)\) to \((2,3)\) lies on \(x+y=5\), so \(x+y\ge5\) is active there. 5. The edge from \((2,3)\) to \((8,0)\) lies on \(x+2y=8\), so \(x+2y\le8\) is active there.

a) The vertices are \((5, 0)\), \((8, 0)\), and \((2, 3)\). b) The edge from \((5,0)\) to \((2,3)\) has \(x+y=5\) active. The edge from \((2,3)\) to \((8,0)\) has \(x+2y=8\) active.

- First interpret the two parallel y-bounds as a strip. - Apply the x-restriction to keep only one side of the vertical boundary. - Substitute the boundary x-value to find the vertical segment’s endpoints.

1. The first two inequalities create the strip \(x-2\le y\le x+1\) between parallel lines, with both boundaries included. 2. The condition \(x\ge0\) keeps the part on or to the right of the y-axis. 3. At \(x=0\), the allowed y-values satisfy \(-2\le y\le1\), so the y-axis boundary segment runs from \((0, -2)\) to \((0, 1)\), including both endpoints. 4. The solution set extends indefinitely to the right, so it is unbounded.

The solution set is the strip \(x-2\le y\le x+1\), restricted to \(x\ge0\). It is unbounded. Its included y-axis segment is from \((0, -2)\) to \((0, 1)\).

- Use the given solution point to choose a side of each boundary line. - Test the point in the expression that defines each edge. - Preserve boundary inclusion in every comparison symbol.

1. The point \((3, 4)\) has \(x<4\), so the selected side of \(x=4\) is \(x\le4\). 2. It has \(y<5\), so the selected side of \(y=5\) is \(y\le5\). 3. It has \(x+y=7>6\), so the selected side of \(x+y=6\) is \(x+y\ge6\). 4. All boundaries are included, so all three comparisons are inclusive.

\(x\le4\), \(y\le5\), and \(x+y\ge6\)

- Translate the total-tray range as two inequalities. - Write a separate condition for the minimum number of cells. - Check the proposed pair against every bound and the count restrictions.

1. The tray-count range is \(x+y\ge10\) and \(x+y\le14\). 2. The cell requirement is \(24x+40y\ge300\). 3. Counts require \(x\ge0\), \(y\ge0\), and whole-number values. 4. For \((8, 4)\), \(8+4=12\), which is between \(10\) and \(14\), and \(24\cdot8+40\cdot4=352\ge300\), so the plan is feasible.

The system is \(x+y\ge10\), \(x+y\le14\), \(24x+40y\ge300\), \(x\ge0\), and \(y\ge0\), with whole-number values. The plan \((8, 4)\) is feasible.

- Restrict the system to one coordinate axis at a time. - Compare the lower and upper bounds produced by the two slanted inequalities. - Check whether the slanted boundaries intersect inside the first quadrant.

1. On the x-axis, \(y=0\). The slanted inequalities become \(x\le6\) and \(x\ge4\), so the boundary segment runs from \((4,0)\) to \((6,0)\). The lower endpoint comes from \(x+2y=4\), and the upper endpoint comes from \(x+y=6\). 2. On the y-axis, \(x=0\). The slanted inequalities become \(y\le6\) and \(2y\ge4\), so the boundary segment runs from \((0,2)\) to \((0,6)\). Again, the lower endpoint comes from \(x+2y=4\), and the upper endpoint comes from \(x+y=6\). 3. The two slanted boundaries intersect at \((8,-2)\), outside the first quadrant, so there are no additional first-quadrant vertices. 4. Therefore, the vertices are \((4,0)\), \((6,0)\), \((0,6)\), and \((0,2)\).

The vertices are \((4,0)\), \((6,0)\), \((0,6)\), and \((0,2)\). On both coordinate axes, \(x+2y=4\) determines the lower endpoint and \(x+y=6\) determines the upper endpoint.

- Solve the two boundary equations together for their shared point. - Restrict the system to one coordinate axis at a time. - Compare the resulting one-variable upper bounds to identify the tighter condition.

1. Solve \(3x+y=15\) and \(x+2y=12\). The intersection is \(\left(\frac{18}{5}, \frac{21}{5}\right)\). 2. On the x-axis, the inequalities give \(x\le5\) and \(x\le12\), so \(3x+y\le15\) is tighter. 3. On the y-axis, the inequalities give \(y\le15\) and \(y\le6\), so \(x+2y\le12\) is tighter. 4. Thus, each slanted boundary controls a different axis side of the solution set.

The boundaries intersect at \(\left(\frac{18}{5}, \frac{21}{5}\right)\). The inequality \(3x+y\le15\) is tighter on the x-axis, and \(x+2y\le12\) is tighter on the y-axis.

- Decide inclusion or exclusion separately for each boundary before looking at the overlap. - Translate \(x+y\le4\) into a statement about points above or below its diagonal boundary. - A corner belongs only if it satisfies every inequality in the system.

1. The inequality \(x\ge0\) includes the y-axis, so \(x=0\) is solid and the satisfying side is to the right. 2. The inequality \(y>0\) excludes the x-axis, so \(y=0\) is dashed and the satisfying side is above. 3. The inequality \(x+y\le4\) includes its boundary \(y=4-x\), so the diagonal is solid and the satisfying side is below it. 4. The common region is the triangular region in panel b). 5. The point \((0,0)\) is excluded because \(y=0\). The point \((0,4)\) satisfies all three inequalities and is included. The point \((4,0)\) is excluded because \(y=0\).

Panel b). The boundaries \(x=0\) and \(x+y=4\) are solid, while \(y=0\) is dashed; the overlap is to the right of the y-axis, above the x-axis, and below the diagonal. \((0,0)\): excluded \((0,4)\): included \((4,0)\): excluded

- Combine the comparison \(x\ge y\) with the lower bound on \(y\). - Remove only a condition that is already guaranteed by the others. - Find intersections of the remaining active boundaries.

1. From \(y\ge0\) and \(x\ge y\), it follows that \(x\ge0\). Therefore, the inequality \(x\ge0\) is redundant. 2. The boundaries \(x=0\) and \(y=0\) meet at \((0, 0)\). 3. On the x-axis, \(x+y=5\) gives \((5, 0)\), which satisfies \(x\ge y\). 4. The boundaries \(x=y\) and \(x+y=5\) give \(2x=5\), so \(x=y=\frac{5}{2}\).

a) \(x\ge0\) is redundant. b) The vertices are \((0, 0)\), \((5, 0)\), and \(\left(\frac{5}{2}, \frac{5}{2}\right)\).

- Express the condition for a lower bound not to exceed an upper bound. - Consider positive and negative x-values separately. - Look for the parameter value that satisfies both resulting requirements.

1. For a fixed x, an overlap requires \(ax\le2x\). 2. When \(x>0\), this requires \(a\le2\). 3. When \(x<0\), division by x reverses the inequality and requires \(a\ge2\). 4. Both requirements hold only when \(a=2\). Then the overlap for every x is the boundary line \(y=2x\).

\(a=2\)

- To show a condition matters, find a point allowed by the other conditions but rejected by that one. - Look near different coordinate axes because the two coefficient patterns favor different directions. - One counterexample is needed for each proposed redundancy.

1. The point \((7, 2)\) is nonnegative and satisfies \(x+y=9\le10\), but \(2x+y=16>12\). Therefore, \(2x+y\le12\) is not redundant. 2. The point \((0, 11)\) is nonnegative and satisfies \(2x+y=11\le12\), but \(x+y=11>10\). Therefore, \(x+y\le10\) is not redundant. 3. Each slanted inequality removes points allowed by the other, so neither is redundant.

Neither slanted inequality is redundant.

- Use the shared boundary point to determine the missing slope. - Test the given solution point against each completed boundary equation. - Use the stated inclusion to choose nonstrict symbols.

1. Substitute \((2, 4)\) into \(y=mx+8\): \(4=2m+8\), so \(m=-2\). 2. The point \((1, 4)\) lies to the right of \(x=0\), so the solution set uses \(x\ge0\). 3. At \((1, 4)\), \(4\ge1+2\), so the solution set is on or above \(y=x+2\). 4. Also, \(4\le-2\cdot1+8\), so the solution set is on or below \(y=-2x+8\).

\(m=-2\). The system is \(x\ge0\), \(y\ge x+2\), and \(y\le-2x+8\).

- Apply the same translation to every boundary and every vertex. - For the slanted boundary, express the old coordinates in terms of the new coordinates. - Check one translated vertex in all three new inequalities.

1. Translating \(x\le0\) right by \(4\) gives \(x\le4\). 2. Translating \(y\ge0\) down by \(1\) gives \(y\ge-1\). 3. Substitute \(x-4\) for x and \(y+1\) for y in \(y\le x+3\): \(y+1\le x-4+3\), so \(y\le x-2\). 4. The original vertices \((-3, 0)\), \((0, 0)\), and \((0, 3)\) translate to \((1, -1)\), \((4, -1)\), and \((4, 2)\).

The translated system is \(x\le4\), \(y\ge-1\), and \(y\le x-2\). Its vertices are \((1, -1)\), \((4, -1)\), and \((4, 2)\).

- Compare the lower and upper boundary expressions directly. - Separate the strict, equality, and reversed-order parameter cases. - Interpret each case geometrically.

1. An overlap requires \(x+b\le x+3\). 2. Canceling x gives \(b\le3\). 3. If \(b<3\), the solution set is the strip between two distinct parallel lines, including both boundaries. 4. If \(b=3\), the solution set is the single line \(y=x+3\). 5. If \(b>3\), the lower boundary lies above the upper boundary, so there are no solutions.

If \(b<3\), the solution set is the strip between the two lines, including both boundaries. If \(b=3\), the solution set is the line \(y=x+3\). If \(b>3\), there are no solutions.

- Determine the line through the two given points. - Use the known solution to choose a side of each boundary. - Because each boundary is included, use nonstrict inequality symbols.

1. The slope through \((1, -1)\) and \((-2, 2)\) is \(\frac{2-(-1)}{-2-1}=-1\), so the third boundary is \(y=-x\), or \(x+y=0\). 2. Since \((0, 1)\) has \(x<1\), the vertical condition is \(x\le1\). 3. Since \((0, 1)\) has \(y<2\), the horizontal condition is \(y\le2\). 4. Since \(0+1>0\), the diagonal condition is \(x+y\ge0\).

The third boundary is \(x+y=0\). The system is \(x\le1\), \(y\le2\), and \(x+y\ge0\).

- Use the two vertices that lie on the unknown slanted edge. - Convert the line through those points to standard form. - Use a known solution to choose each inequality direction.

1. The unknown slanted boundary passes through \((0,4)\) and \(\left(\frac{18}{5},\frac{14}{5}\right)\). 2. Its slope is \(\frac{\frac{14}{5}-4}{\frac{18}{5}-0}=-\frac{1}{3}\), so its equation is \(y=-\frac{1}{3}x+4\), or \(x+3y=12\). 3. The origin is a solution, so the selected sides are \(x+3y\le12\) and \(2x+y\le10\). 4. The first-quadrant restrictions are \(x\ge0\) and \(y\ge0\).

The other slanted boundary is \(x+3y=12\). The complete system is \(x\ge0\), \(y\ge0\), \(x+3y\le12\), and \(2x+y\le10\).

- Translate all three requirements before searching for whole-number pairs. - Fix the number of live demonstrations and find the allowable range for the recorded segments. - Stop when the minimum segment requirement exceeds what the time limit allows.

1. The constraints are \(x+y\ge7\), \(x\ge2\), \(20x+8y\le100\), with nonnegative whole-number values. 2. If \(x=2\), the count condition gives \(y\ge5\), and the time condition gives \(y\le7\), producing \((2, 5)\), \((2, 6)\), and \((2, 7)\). 3. If \(x=3\), the count condition gives \(y\ge4\), and the time condition gives \(y\le5\), producing \((3, 4)\) and \((3, 5)\). 4. If \(x=4\), the count condition gives \(y\ge3\), but the time condition gives \(y\le2\), so no pair works. Larger x-values also cannot work. 5. Therefore, there are five feasible ordered pairs.

The feasible ordered pairs are \((2, 5)\), \((2, 6)\), \((2, 7)\), \((3, 4)\), and \((3, 5)\).