Graphical solution of linear inequalities in two variables

The direct answer is option B, 4 vertices. Begin with the rectangle in the first quadrant: 0≤x≤5 and 0≤y≤3. The condition x+y≥4 keeps the part on or above the line x+y=4. The line meets the top edge y=3 at x=1, giving (1,3), and it meets the bottom edge y=0 at x=4, giving (4,0). The remaining rectangle corners (5,3) and (5,0) also satisfy x+y≥4. Thus the four vertices are (1,3), (5,3), (5,0), and (4,0). Option A, 3, misses one corner. Option B, 4, counts all four boundary turns. Option C, 5, incorrectly counts an intersection that is not a vertex or counts an excluded rectangle corner. Option D, 6, overcounts points along an edge. Remember that a vertex is a corner where the boundary changes direction, not every point on a line.

In option A, the region lies in the first quadrant and is enclosed by \(x+y=4\). Its vertices are \((0,0),(4,0),(0,4)\), so it is a bounded triangle. In exams, check whether the shaded region extends infinitely in any direction.

Solving the two equations gives \(x=\frac{27}{7}\) and \(y=\frac{32}{7}\). The intersection should then be checked in the inequalities.

Substituting the origin gives (0<10), which is true. Because the inequality is strict, the boundary is dashed and shading is toward the origin.

The direct answer is option A, 6. We want x to be as large as possible. The constraint y≥1 says the smallest allowed y is 1. Since x+2y≤8, replacing y by its smallest value gives x+2(1)≤8, so x+2≤8 and x≤6. This bound is attainable at (6,1), which satisfies x≥1, y≥1, and x+2y=8. Hence the maximum is 6. Option A, 6, works exactly at the boundary. Option B, 7, would give 7+2=9>8, so it is not feasible. Option C, 8, would give 10>8 and is also impossible. Option D, 5, is feasible, for example with y=1, but it is not the maximum because 6 is feasible. The key idea is that increasing y consumes two units of the allowance, so for maximum x we choose the smallest permitted y. Exam cue: for a positive coefficient of y in an upper-bound inequality, minimize y when maximizing x.

The direct answer is option C, 0≤y≤4. From x≥0 and x+3y≤12, we have 3y≤12, so y≤4. Together with y≥0, this gives 0≤y≤4. Both endpoints are attainable: when y=0, choose x=4, which satisfies x+y≥4 and x+3y≤12; when y=4, choose x=0, which satisfies x+y=4 and x+3y=12. Thus the range is closed, including 0 and 4. Option A, 1≤y≤4, incorrectly excludes the feasible point y=0. Option B, 0<y<4, incorrectly excludes both feasible boundary values. Option C, 0≤y≤4, includes exactly every possible value. Option D, 2≤y≤6, has an incorrect lower bound and exceeds the upper bound. A useful method is to use x≥0 to obtain the largest possible y, then check whether the lower endpoint is allowed by the other conditions. Memory cue: inequalities with ≤ and ≥ usually allow their boundary values.

The same expression (x+y) cannot be at most (3) and at least (5) at the same time. Identify such contradictions before graphing.

If (a\geq 0), at least ((0,0)) is in the solution and the region remains bounded. If (a<0), there is no solution in the first quadrant.

The vertices are ((1,0)), ((4,0)), ((3,2)), and ((1,3)). The shoelace method gives area (6) square units.

At ((5,1)), (x-y=4), which is greater than (2). In option testing, one failed inequality excludes the point.

Checking (x+2y) at the corners gives the maximum (12) at ((2,5)). For a linear expression, checking corners is sufficient.

The three strict inequalities form a bounded triangle, but no boundary is included. Therefore the region is open and bounded.

Corner points are intersections of boundary lines that satisfy every inequality. The feasible vertices are \((0,0)\), \((6,0)\), \((4,2)\), and \((0,4)\). Therefore, the solution region has 4 corner points. For example, \((0,6)\) is not feasible because it does not satisfy \(x+2y\leq 8\). Exam tip: Test every candidate intersection in all the inequalities.

A solid boundary includes equality. Since the origin gives (0<12), the side opposite the origin is (3x+2y\geq 12).

Right side including the line means the value of (x) is greater than or equal to (-2). For vertical lines, read the (x)-value directly.

The line \(y=3\) divides the plane into points above it and points below it. “Below” means that the y-coordinate is smaller than 3, so the basic inequality is \(y<3\). The phrase “without including the line” means equality is not allowed. Consequently, the required inequality is \(y<3\), which is choice D.

If the line itself were included, the correct sign would be \(\leq\), because points with y=3 would also belong to the region. The sign \(y>3\) describes the region above the line, while \(y\geq3\) includes the line and also describes the upper side. On a graph, \(y=3\) is drawn as a horizontal dashed line for the strict inequality, and the region below it is shaded.

To maximize (y), take the minimum allowed (x=1). Then (2+y\leq 9) and (1+y\leq 7) give (y\leq 6).

To obtain the minimum value of \(y\), take the largest allowed value of \(x\), namely \(4\). Then \(x+y\geq 6\) gives \(y\geq 2\). Also, substituting \(x=4\) in \(2x+y\geq 7\) gives \(y\geq -1\), while \(y\geq 0\) is already given. Combining all the lower bounds gives \(y\geq 2\), so the minimum value is \(2\). The point \((4,2)\) satisfies every inequality. Exam tip: when minimizing a variable, compare all lower bounds imposed by the constraints.

Both boundary lines have slope 2, so they are parallel. For every value of \(x\), \(2x+3\) is 4 greater than \(2x-1\). Thus, no value of \(y\) can be simultaneously less than or equal to \(2x-1\) and greater than or equal to \(2x+3\). Hence, there is no common region. Exam tip: for parallel boundary lines, compare the constant terms to quickly check whether the half-planes overlap.

The first two conditions give (x+y=6), and the next two give (x-y=2). Their intersection is ((4,2)).

The inequalities x≥0 and y≥0 restrict the points to the first quadrant, including the axes. For each integer x from 0 through 4, the condition x+y≤4 gives y=0 through 4-x. Thus the numbers of possible integer y-values are 5 when x=0, 4 when x=1, 3 when x=2, 2 when x=3, and 1 when x=4. Adding them gives 5+4+3+2+1=15. Equivalently, these points form a triangular lattice under x+y=4, including all boundary points. Therefore option B is correct. Counting only interior points or excluding the axes would produce smaller, incorrect totals.

Since \(x\geq 1\) and \(x+y\leq 5\), the possible integer values of \(x\) are \(1,2,3,4\). For these values, the allowable positive integer values of \(y\) are respectively \(1\) to \(4\), \(1\) to \(3\), \(1\) to \(2\), and \(1\) to \(1\). Therefore, the number of points is \(4+3+2+1=10\). An option such as \(8\) misses some valid boundary points. Exam tip: when an inequality includes equality, count the points on the boundary line as well.

The square has area (25), and the triangle below (x+y=4) has area (8). Therefore the remaining area is (17).

Since \(x\geq 0\) and \(y\geq 5\), it must follow that \(x+y\geq 5\). However, the third inequality requires \(x+y\leq 4\), which is incompatible with this condition. Therefore, no point satisfies all three inequalities simultaneously, so the solution region is empty. Even on the boundary line \(x+y=4\), \(y\geq 5\) is impossible when \(x\geq 0\). Exam tip: Compare minimum and maximum possible values to spot an inconsistency quickly.

The two half-planes intersect to form an infinite angular region. Since equality is included in both, the region is closed.