Graphical solution of linear inequalities in two variables

Each linear inequality represents a half-plane, and the intersection of half-planes is convex. A weighted average of any two solutions also satisfies every inequality. Exam tip: test boundedness and feasibility separately.

Substitution gives (2(4)+3(2)=14), which is greater than (12). In exams, decide the half-plane first using a test point.

For a linear inequality, first draw the corresponding boundary equation by replacing the inequality sign with an equality. Here the boundary is \(3x-2y=6\). Because the original sign is “less than,” points exactly on this line do not satisfy the inequality and must not be included in the solution set. Therefore the line is drawn dashed, so choice B is correct.

A solid line is used for inclusive signs such as \(\leq\) or \(\geq\), because those include boundary points. A dashed line is used for strict signs such as \(<\) or \(>\). After drawing the dashed boundary through suitable intercepts, a test point can identify which side satisfies \(3x-2y<6\). The axes alone are not the boundary because the equation is neither \(x=0\) nor \(y=0\).

Adding the two equations gives (2x=12), so (x=6) and (y=4). To find a vertex, solve the two boundary lines together.

Solving both lines together gives (x=4), (y=4). The intersection of shared boundaries is often the key vertex.

Both boundaries are parallel, and the region lies between them. For inequalities with the same left side, compare the constants.

The same quantity (x+y) cannot be at most (4) and at least (7) at the same time. Identify contradictory inequalities first.

Rewriting gives \(y<\frac{2}{3}x-2\), so the required region lies below the boundary line. Since the inequality is strict, the boundary is dashed, not solid. Exam tip: reverse the inequality sign when dividing by a negative number.

The lines (x=2) and (y=1) meet at ((2,1)), and it also satisfies (x+y\le 8). Check every vertex in all inequalities.

The region above the line in the first quadrant extends indefinitely. Conditions with (\ge) often create outward unbounded regions.

The rectangle has length (5-1=4) and width (6-2=4), so the area is (16). For parallel bounds, take differences to get dimensions.

For any point on the \(y\)-axis, \(x=0\). Substituting \(x=0\) in the boundary line \(2x+y=10\) gives \(y=10\), so the corner is \((0,10)\). Although \((0,5)\) satisfies the inequality, it is not a vertex formed by the boundary line and the axes. Exam tip: to find an axis intercept, set the other variable equal to zero.

At ((8,0)), (3(8)+4(0)=24), so it is on the boundary and included by (\le). For (\le) and (\ge), the boundary line is solid.

Rearranging gives (y\ge \frac{5x-10}{2}). When dividing by a negative coefficient, the inequality sign reverses.

At ((2,3)), (x+y=5), so it is not on (x+y=6). To test a boundary, substitute the point into the equation.

Among the options, ((5,4)) satisfies both equations (2x+y=14) and (x+3y=18). While checking options, substitute into both lines.

At ((6,1)), (x+2y=8) and (x+y=7\le 10), so it lies on the second boundary. To identify a boundary point, check equality.

The region below gives smaller values, and including the boundary requires (\le). Use a solid line to include equality.

The intercepts are ((9,0)) and ((0,3)), so area is (\frac{1}{2}\times 9\times 3=\frac{27}{2}). For an axial triangle, use intercepts as base and height.

The upper bound (2x+3y\le 18) and the first quadrant make the region bounded. The lower line (x+y\ge 4) only cuts an inner part.

For ((1,4)), (4\le 4) and (4\ge 0), so both inequalities hold. Both conditions must be checked together.

(-5y\ge 15-2x), and dividing by (-5) reverses the sign to (y\le \frac{2x-15}{5}). Dividing by a negative number reverses the inequality sign.

Solving \(x+2y=10\) and \(3x+y=12\) gives \(\left(\frac{14}{5},\frac{18}{5}\right)\). Fractional vertices are also valid in graphs.

There are upper and right bounds, but the region can extend indefinitely toward the lower-left direction. To test boundedness, check restrictions in every direction.

Together (x+y\le 6) and (x+y\ge 6) force (x+y=6). Equality can be written using two opposite inequalities.