Graphical solution of linear inequation

This sub-section works through six fully solved examples applying the shading method of section 7.1.2.

Example 1 — six single inequations, each solved by finding the boundary line and testing a point.

a) $x\le3$: the boundary is the vertical line $x=3$. The origin $(0,0)$ gives $0\le3$, true, so the origin side (the

region to the LEFT of $x=3$, including the line) is shaded.

b) $y\ge-2$: the boundary is the horizontal line $y=-2$. The origin gives $0\ge-2$, true, so the origin side (the region

ABOVE $y=-2$, including the line) is shaded.

c) $x+2y\le0$: rewritten as $x=-2y$, this line passes through the origin, so the origin cannot serve as the test point.

Choosing $(-2,-2)$ instead: $x+2y=-2+2(-2)=-2-4=-6$, and $-6\le0$ is true, so the half-plane containing $(-2,-2)$ is

shaded.

d) $2x+3y\ge6$: the boundary line passes through $(3,0)$ and $(0,2)$ (its intercepts). Testing the origin:

$2(0)+3(0)=0$, and $0\ge6$ is FALSE, so the origin side is excluded — the required region is the NON-origin side of the

line.

e) $2x-3y\ge-6$: the boundary line passes through $(-3,0)$ and $(0,2)$. Testing the origin: $2(0)-3(0)=0$, and

$0\ge-6$ is TRUE, so the required region IS the origin side of the line.

f) $4x-5y\le20$: the boundary line passes through $(5,0)$ and $(0,-4)$. Testing the origin: $4(0)-5(0)=0$, and

$0\le20$ is TRUE, so the required region is the origin side of the line.

Example 2 — $3x+2y\le6$. Draw the line $3x+2y=6$ through its intercepts $(2,0)$ and $(0,3)$. Testing the origin:

$3(0)+2(0)=0\le6$, true — the origin side of the line is the required region.

Example 3 — the system $x+2y\ge4$ and $2x-y\le6$. Draw both boundary lines: $x+2y=4$ through $(4,0)$ and $(0,2)$, and

$2x-y=6$ through $(3,0)$ and $(0,-6)$. Testing the origin in each: for $x+2y\ge4$, $(0,0)$ gives $0\ge4$, FALSE, so this

constraint's shaded region is the NON-origin side; for $2x-y\le6$, $(0,0)$ gives $0\le6$, TRUE, so this constraint's

shaded region is the origin side. The common (overlapping) region of these two oppositely-labelled half-planes is the

graphical solution of the system — an unbounded region (since neither inequation caps the plane on every side), whose

single corner point is found by solving $x+2y=4$ and $2x-y=6$ simultaneously: from the second equation $y=2x-6$;

substituting into the first, $x+2(2x-6)=4 \Rightarrow x+4x-12=4 \Rightarrow 5x=16 \Rightarrow x=3.2,\ y=0.4$. So the

corner point is $(3.2,\,0.4)$.

Example 4 — the system $3x+4y\le12$ and $x-4y\le4$. Draw both lines: $3x+4y=12$ through $(4,0)$ and $(0,3)$, and

$x-4y=4$ through $(4,0)$ and $(0,-1)$ (note both lines happen to pass through the same point $(4,0)$). Testing the

origin in each gives $0\le12$ (true) and $0\le4$ (true, in fact $0<4$ strictly), so BOTH constraints shade their origin

side, and the common shaded region — the graphical solution — is the overlap of the two origin-side half-planes, an

unbounded wedge-shaped region with its one finite corner at the shared point $(4,0)$ where the two lines cross.

Note on plotting a line via the double-intercept form. As an alternative to picking two arbitrary points, any line

$ax+by=c$ (with $c\ne0$) can be written in double-intercept form by dividing through by $c$: for example $3x+2y=6$

becomes $\dfrac{x}{2}+\dfrac{y}{3}=1$, immediately showing the $x$-intercept is $2$ and the $y$-intercept is $3$, i.e. the

line passes through $(2,0)$ and $(0,3)$ — the same two points used to draw it above.

Feasible solutions of a system of inequations (introducing the term). When several linear inequations must hold at the same time — together with the non-negativity constraints $x\ge0,\ y\ge0$ — the region common to all of them is called the feasible region, and every point of it is a feasible solution. The same origin-side / non-origin-side shading method of section 7.1.2 is applied to each constraint, and the overlap of all the shaded half-planes is the feasible region.

Example 1. Find the graphical solution of the system $2x+y\le10,\ 2x-y\le2,\ x\ge0,\ y\ge0$. Draw $L_1:2x+y=10$ (through $(0,10)$ and $(5,0)$) and $L_2:2x-y=2$ (through $(0,-2)$ and $(1,0)$); testing the origin, each constraint shades its origin side, and within the first quadrant the common shaded region $OABCO$ is a bounded quadrilateral with vertices $O(0,0)$, $A(1,0)$, $B(3,4)$ and $C(0,10)$. This common shaded region is the feasible solution of the system.

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