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 equation (I), \(2+1=3<4\), so the point satisfies the inequality strictly. 2. For equation (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 equation (II) by \(2\) to get \(x+y\le7\). 2. Every point satisfying equation (I), \(x+y\le6\), also satisfies \(x+y\le7\). 3. Therefore, equation (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 equation (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.

The student 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 \(c\).

- 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 equation (I) provides the tighter upper bound. 4. For \(x>-1\), \(-2x+1<x+4\), so equation (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.

- Identify the boundary equations in each graph. - Use vertical and horizontal restrictions before comparing slanted boundaries. - A system’s graph is the overlap of all its inequalities.

1. Graph 1 is left of \(x=2\) and above \(y=-1\), so it matches system C. 2. Graph 2 is in the first quadrant and below \(x+y=4\), so it matches system A. 3. Graph 3 is the strip between the parallel lines \(y=x-1\) and \(y=x+2\), so it matches system B.

A) Graph 2 B) Graph 3 C) Graph 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 \(x<2\).

- 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 the intersections of boundary lines that satisfy every inequality. - Compare squared distances \(x^2+y^2\) instead of taking every square root. - The largest distance identifies the farthest vertex.

1. The axes meet at \((0, 0)\). 2. On the x-axis, \(x+y=6\) gives \((6, 0)\). 3. The lines \(y=4\) and \(x+y=6\) intersect at \((2, 4)\). 4. The line \(y=4\) meets the y-axis at \((0, 4)\). 5. The distances from the origin are \(0\), \(6\), \(\sqrt{20}\), and \(4\), respectively, so \((6, 0)\) is farthest.

The vertices are \((0, 0)\), \((6, 0)\), \((2, 4)\), and \((0, 4)\). The farthest vertex from the origin is \((6, 0)\).

- 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. - Use the horizontal base and vertical height of the triangle.

1. The line \(y=2x+4\) meets the x-axis at \((-2, 0)\). 2. The line \(y=-x+7\) meets the x-axis at \((7, 0)\). 3. The two slanted boundaries intersect where \(2x+4=-x+7\). This gives \(x=1\) and \(y=6\). 4. The triangle has base length \(7-(-2)=9\) and height \(6\), so its area is \(\frac{1}{2}\cdot9\cdot6=27\) square units.

The vertices are \((-2, 0)\), \((7, 0)\), and \((1, 6)\). The area is \(27\) square units.

- 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.

- Find where each slanted boundary meets each coordinate axis. - Check whether the two slanted lines intersect in the first quadrant. - Subtract endpoint coordinates to find the lengths of the axis-aligned boundary segments.

1. On the x-axis, \(x+2y=4\) gives \((4, 0)\), and \(x+y=6\) gives \((6, 0)\). 2. On the y-axis, \(x+2y=4\) gives \((0, 2)\), and \(x+y=6\) gives \((0, 6)\). 3. The two slanted boundary lines intersect at \((8, -2)\), which is outside the first quadrant. Therefore, the vertices are \((4, 0)\), \((6, 0)\), \((0, 6)\), and \((0, 2)\). 4. The x-axis boundary segment has length \(6-4=2\) units. The y-axis boundary segment has length \(6-2=4\) units, so the y-axis segment is \(2\) units longer.

The vertices are \((4, 0)\), \((6, 0)\), \((0, 6)\), and \((0, 2)\). The x-axis boundary segment is \(2\) units long, and the y-axis boundary segment is \(4\) units long, so the y-axis segment is \(2\) units longer.

- 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.

- On each axis, use the stricter resource limit to find the endpoint. - Solve the two slanted boundary equations together for the non-axis vertex. - Order the vertices around the boundary before using an area method.

1. The coordinate axes meet at \((0, 0)\). 2. On the x-axis, the tighter upper limit is \(x\le6\), giving \((6, 0)\). 3. The two slanted boundaries intersect where \(x+y=6\) and \(x+2y=8\). Subtraction gives \(y=2\), then \(x=4\), so the intersection is \((4, 2)\). 4. On the y-axis, the tighter upper limit is \(2y\le8\), giving \((0, 4)\). 5. Using the vertices in order \((0, 0)\), \((6, 0)\), \((4, 2)\), and \((0, 4)\), the shoelace formula gives area \(\frac{1}{2}(28)=14\) square units.

The vertices are \((0, 0)\), \((6, 0)\), \((4, 2)\), and \((0, 4)\). The area is \(14\) square units.

- Find all valid intersections of the boundary lines. - A linear expression on a bounded polygon reaches a maximum at a vertex. - Evaluate the expression at each vertex and compare the results.

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 \(x+2y=8\). Subtraction gives \(y=3\), then \(x=2\), so the intersection 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. Evaluate \(x+4y\) at the vertices: \(5\), \(8\), and \(14\). The maximum is \(14\) at \((2, 3)\).

a) The vertices are \((5, 0)\), \((8, 0)\), and \((2, 3)\). b) The maximum value is \(14\), attained at \((2, 3)\).

- 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 \(\begin{cases}x\le1\\y\le2\\x+y\ge0\end{cases}\).

- 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 \(\begin{cases}x\ge0\\y\ge0\\x+3y\le12\\2x+y\le10\end{cases}\).

- 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 \((3, 5)\).