Graphing Inequalities: Core practice

Adding $y$ gives

Dividing by positive 2 gives

The second coordinate is free.

Draw a dashed vertical line halfway between 1 and 2 and shade its left side.

The origin gives $0<3$, true.

The point $(2,0)$ gives $4<3$, false, confirming the chosen side.

Substitution gives

Subtracting 4 from both sides and dividing by 2 gives

With this coefficient, the boundary is $x+y=6$, and $2+4=6$ confirms the point lies on it.

The solid boundaries are $x=-3$, $y=0$, and $x+3y=6$.

Keep points right of the first, above the second, and below the third.

Their pairwise crossings that satisfy every condition are $(-3,0)$, $(6,0)$, and $(-3,3)$.

Shade this triangle including its edges.

On $x=3$, the third inequality becomes

Subtracting 3 from both sides and dividing by 3 gives $y\le1$, and $y\ge0$ still applies.

The slice is the segment from $(3,0)$ to $(3,1)$ with both endpoints included, each of which satisfies all three conditions.

All four boundaries are solid.

The vertical boundary gives corners $(0,1)$ and $(0,6)$.

The lower horizontal boundary meets $y=x-2$ at $(3,1)$.

The two slanted boundaries meet where

Then $y=2$, giving $(4,2)$.

Shade the quadrilateral inside these four edges.

Each listed corner satisfies every constraint.

At $(3,2)$, the four checks are $3\ge0$, $2\ge1$, $5\le6$, and $2\ge3-2$, so the point belongs.

A point is visible when it is on the screen and fails $x+2y\ge8$, that is when

The symbol is strict, so this cut is dashed and the points on it are hidden.

The origin gives $0<8$, so keep its side.

The cut meets $x=6$ where $y=1$, and meets $y=4$ where $x=0$.

Shade the quadrilateral with vertices $(0,0)$, $(6,0)$, $(6,1)$, and $(0,4)$, leaving the dashed edge from $(0,4)$ to $(6,1)$ out, so those two vertices are not visible themselves.

Testing the four screen corners, $(0,0)$ gives 0 and $(6,0)$ gives 6, both under 8, while $(6,4)$ gives 14 and $(0,4)$ gives exactly 8.

So the filter hides $(6,4)$ and $(0,4)$.

With $x=3$, the first condition becomes

Adding $y$ to both sides and then subtracting 3 gives

Combined with $y<7$, the allowed heights are $3\le y<7$.

The smallest upward move from height 0 reaches height 3, a move of 3 units.

At $(3,3)$ the first condition gives $2(3)-3=3$, which sits exactly on the boundary $2x-y=3$, and the inclusive symbol counts that boundary as part of the region; the second gives $3<7$.

A shorter upward move would leave $y<3$ and fail the first condition.

A single linear inequality selects one side of its boundary.

For a nonvertical boundary, points above it and below it give opposite signs for their height difference from the line.

A vertical boundary separates smaller and larger first coordinates in the same way.

The two points lie strictly on different sides, so exactly one lies on the selected side.

Changing between strict and inclusive symbols affects the boundary itself, not either of these off-boundary points.

Any point satisfying the enlarged list satisfies every inequality in the original list.

Therefore it lies in the original feasible region.

The new condition can remove old points or leave the region unchanged, but it cannot make a previously failing original condition true.

Thus no point outside the old region can enter the new one.

The point $(0,3)$ satisfies $0\le2$ but fails $3\le2$, so it belongs to A and not B.

The point $(3,0)$ fails A but satisfies B.

These two points disprove both possible containments.

A is the half-plane on or left of a vertical boundary; B is the half-plane on or below a horizontal boundary.