Graphical solution of linear inequalities in two variables

Writing the line as (y=\frac{18-2x}{3}), the upper side is (y\ge\frac{18-2x}{3}). This is equivalent to (2x+3y\ge18).

The conditions (0\le x\le6) and (0\le y\le5) restrict the region inside a rectangle. The inequality (x+y\ge4) only cuts off one corner.

Solving \(x+2y=9\) and \(3x+y=11\) gives \(x=\frac{13}{5}\), \(y=\frac{16}{5}\). Fractional vertices are common in graphical solutions.

Both lines have slope (3), so they are parallel. The inequalities give a strip-like region between them.

The intercepts are ( (6,0) ) and ( (0,4) ), so the area is (\frac{1}{2}\times6\times4=12). For a triangular region, identify base and height.

The (x)-axis is (y=0), and below it (y) is less than or equal to zero. Use (\le) when the boundary is included.

The boundaries (x=1), (y=2), and (x+y=8) give ( (1,2) ), ( (6,2) ), and ( (1,7) ). Do not treat shifted restrictions as simple (x\ge0), (y\ge0).

It is impossible for (x+y) to be both less than and greater than (5) at the same time. Opposite inequalities can make the solution empty.

The vertices are ( (0,0) ), ( (3,0) ), ( (2,2) ), and ( (0,3) ), so every listed point is a vertex. The correct choice should have been none of these.

From (x-3y\le9), we get (-3y\le9-x), hence (y\ge\frac{x-9}{3}). Dividing by a negative reverses the sign.

At ( (2,2) ), both left sides equal (10), which is less than (12). Therefore this option cannot be correct and true solutions must make both values at least (12).

On the boundary line \(x+2y=10\), put \(x=2\). Then \(2+2y=10\), so \(y=4\) and one vertex is \((2,4)\). Similarly, putting \(y=1\) gives \(x+2=10\), so \(x=8\) and the other vertex is \((8,1)\). The points in option D lie on the line, but they are not vertices formed by intersections with the boundary lines \(x=2\) and \(y=1\). Exam tip: find vertices by checking intersections of pairs of boundary lines.

Adding the equations gives \(4x+4y=30\), and by symmetry \(x=y=\frac{15}{4}\). Use symmetry when boundary lines have a similar structure.

The value (x+y) cannot be at most (4) and at least (7) simultaneously. Parallel bounds in opposite positions can give an empty region.

At ( (1,4) ), (1\le3) and (4\ge3) are true. Check both the vertical boundary and the oblique boundary for each option.

The two equations give \(x=y\) and \(3x=14\), so the vertex is \( \left(\frac{14}{3},\frac{14}{3}\right) \). Symmetric boundaries can quickly give \(x=y\).

An inequality in the form (y\le) represents the region below the line. The equality part means the boundary line is included.

Both inequalities select regions above the boundary lines, and the common region extends indefinitely. Conditions with (\ge) can often create an unbounded region.

Left of (x=4) is (x\le4), and above (y=-1) is (y\ge-1). For vertical and horizontal boundaries, identify direction separately.

At ( (1,1) ), (x+y=2), which is less than (3). In multi-inequality questions, one false condition is enough to reject a point.

At (y=1), (x\le8) and (x\le4), so the tighter limit is (x=4). For the right endpoint, choose the smaller upper bound from all constraints.

At ( (3,3) ), (6+9=15<18), so it is inside and not on the axes. For an interior point, look for a strict inequality value.

Two vertical and two horizontal boundaries together form a closed rectangle. In a graph, pairs of parallel boundaries help identify the shape quickly.

From (x+2=-x+4), (2x=2), so (x=1) and (y=3). The intersection of slant boundaries can mark the start or a corner of a region.

Putting the origin gives (0\leq 6), so the side containing it is taken. Exam tip: (\leq) includes the boundary line.