Inequalities: Chapter Test

The filled circle means the boundary $-3$ belongs to the set, so it takes a bracket, and the shading runs toward the larger numbers, so the set is unbounded above. The infinite end always takes a parenthesis.

$x \ge -3 \quad\Longleftrightarrow\quad [-3, \infty)$

Add $8$ to both sides, which leaves the direction alone, then divide by the positive number $5$, which also leaves it alone.

$5x - 8 \le 7 \quad\Longrightarrow\quad 5x \le 15 \quad\Longrightarrow\quad x \le 3$

The symbol admits equality, so every point of the line is a solution and the boundary is drawn solid. For the side, substitute a point that is not on the line; the origin is the cheapest.

$0 \le -\tfrac{1}{2}(0) + 4 \quad\Longrightarrow\quad 0 \le 4$

That is true, so the origin is a solution and its whole side is shaded.

Subtracting the same amount from both sides slides them equally along the line, so the order survives. Dividing by a negative number reflects both sides across zero, which swaps their order, so the symbol must reverse there.

$9 - 4x > 1 \quad\Longrightarrow\quad -4x > -8 \quad\Longrightarrow\quad x < 2$

Each end chooses its own fence from its own symbol. The left bound is strict, so $-5$ is excluded and takes a parenthesis; the right bound admits equality, so $-1$ is included and takes a bracket.

$-5 < x \le -1 \quad\Longleftrightarrow\quad (-5, -1]$

The objective is linear and the region is bounded, so its largest value over the region is reached at a corner. Evaluate it at each of the four.

$P(0,0) = 0, \qquad P(6,0) = 12, \qquad P(4,3) = 8 + 18 = 26, \qquad P(0,5) = 30$

The largest of the four is $30$, at $(0, 5)$. The heavy weight on $y$ is what lifts the tall corner past the wide one.

Read each ray on its own. The filled circle on $1$ with shading to the left is every number at most $1$; the hollow circle on $4$ with shading to the right is every number above $4$. A point is shaded when it lies in one ray or the other, which is a union.

$x \le 1 \;\text{ or }\; x > 4 \quad\Longleftrightarrow\quad (-\infty, 1] \cup (4, \infty)$

The two pieces do not overlap and do not meet, so no single chain can describe them.

Collect the variable terms on the side that leaves their coefficient positive, so that the final division needs no reversal. Adding $3x$ to both sides puts them on the right.

$4 - 3x > 2x + 19 \quad\Longrightarrow\quad 4 > 5x + 19 \quad\Longrightarrow\quad -15 > 5x \quad\Longrightarrow\quad -3 > x$

Read from the other end, that is $x < -3$.

Substitute each pair into the left-hand side and judge the statement it produces.

$(1,1): \; 3 - 4 = -1 \;\text{and}\; -1 < 6 \quad\checkmark \qquad (2,0): \; 6 - 0 = 6 \;\text{and}\; 6 < 6 \quad\times$

The point $(5, 2)$ gives $15 - 8 = 7$ and $(0, -2)$ gives $0 + 8 = 8$, both of which fail. The near miss is $(2, 0)$: it sits exactly on the boundary $3x - 4y = 6$, and a strict symbol keeps the boundary out of the solution set, so an equality there is a failure rather than a pass.

A corner is where two boundaries cross AND every remaining constraint still holds. The boundaries $x = 2$ and $x + y = 7$ cross where

$2 + y = 7 \quad\Longrightarrow\quad y = 5$

so the crossing is $(2, 5)$, and it satisfies $x \ge 0$ and $y \ge 0$ as well. The crossing of $y = 0$ with $x + y = 7$ is $(7, 0)$, which fails $x \le 2$ and is therefore no corner of this region at all.

Work on all three parts at once. Subtract $3$ from each, then divide each by $-2$; that divisor is negative, so both symbols reverse and the two bounds change ends.

$-5 \le 3 - 2x < 7 \quad\Longrightarrow\quad -8 \le -2x < 4 \quad\Longrightarrow\quad 4 \ge x > -2$

Rewritten with the smaller bound first, that is $-2 < x \le 4$. Each fence travels with its own bound: the strict end of the original became the lower bound, so $-2$ is excluded, and the inclusive end became the upper bound, so $4$ is admitted.

Let $b$ be the number of boxes. The total weight is the operator plus the boxes, and it may reach the rating but not pass it.

$170 + 60b \le 2000 \quad\Longrightarrow\quad 60b \le 1830 \quad\Longrightarrow\quad b \le 30.5$

The boundary $30.5$ is not a possible number of boxes, and the admissible whole numbers are the ones at or below it, so the largest is $30$. Checking it: $170 + 60(30) = 1970$ pounds, which is within the rating, while $31$ boxes would give $2{,}030$.

Take one boundary at a time. The horizontal line is solid, so its points belong, and the shading sits below it, giving $y \le 3$. The slanted line climbs one unit for each unit to the right, so it is $y = x$; it is dashed, so its points are excluded, and the shading sits above it, giving $y > x$. A point inside the wedge confirms both:

$(-3, 0): \quad 0 \le 3 \;\text{ and }\; 0 > -3$

Both statements are true, so $(-3, 0)$ belongs to the system, as the picture shows.

The direction of an inequality survives a shift, and it survives scaling by a positive number; scaling by a negative number reverses it. So a multiplier decides which of two outcomes is correct, and that decision is made by its sign alone. With the sign of $t$ unknown, both outcomes are still live:

$t = 2: \; 2a < 2b, \qquad t = -2: \; -2a > -2b$

So the move is not that multiplying by a variable is forbidden; it is that you cannot tell whether to reverse the symbol until the sign is settled.

The region has two corners. The boundaries $x = 0$ and $x + y = 5$ meet at $(0, 5)$, and $y = 1$ and $x + y = 5$ meet at $(4, 1)$; the crossing of $x = 0$ with $y = 1$ is $(0, 1)$, which fails $x + y \ge 5$ and so is no corner here.

$P(0, 5) = 15, \qquad P(4, 1) = 16 + 3 = 19$

From any feasible point, increasing $x$ or increasing $y$ keeps every constraint satisfied and raises $P$ without limit, so there is no maximum. Every direction the region runs on raises $P$, so the smallest value is reached at a corner, and it is $15$.

The symbol admits its boundary in either setting, so the circle is filled and the line is solid. The solutions are the values of $x$ from $-1$ upward, which on a line is the ray to the right and in the plane is every point whose first coordinate is at least $-1$, a vertical boundary with the region to its right. A test point settles the plane picture:

$(2, -7): \quad 2 \ge -1 \;\text{ is true}$

The second coordinate never enters the condition, which is exactly why the boundary runs vertically and the shading is decided by $x$ alone.

Distribute first, then collect. The variable terms are identical on the two sides, so they cancel together and leave a statement with no $x$ in it.

$5 - 3(x + 2) \ge 4 - 3x \quad\Longrightarrow\quad -1 - 3x \ge 4 - 3x \quad\Longrightarrow\quad -1 \ge 4$

What is left decides the whole question, and $-1 \ge 4$ is false. No value of $x$ can repair it, so the solution set is empty. Had the leftover been true instead, every real number would have qualified.

First list the corners, keeping only crossings that satisfy every remaining constraint. The origin is one; $y = 0$ meets $x = 4$ at $(4, 0)$; $x = 4$ meets $2x + y = 10$ at $(4, 2)$; and $x = 0$ meets $2x + y = 10$ at $(0, 10)$. The crossing of $y = 0$ with $2x + y = 10$ is $(5, 0)$, which fails $x \le 4$ and is discarded.

$P(0,0) = 0, \qquad P(4,0) = 16, \qquad P(4,2) = 18, \qquad P(0,10) = 10$

The largest is $18$, at $(4, 2)$. Note that the discarded point would have given $20$, which is exactly why a crossing has to be tested against the other constraints before it counts.

Multiplying both sides by a negative number reverses the symbol, and doing both together leaves the truth of the statement untouched for every pair $(x, y)$. So the two inequalities have identical solution sets, and identical solution sets are identical graphs.

$(0, 3): \; -0 + 6 > 4 \;\text{ is true}, \qquad 0 - 6 < -4 \;\text{ is true}$

The boundaries agree too, since $-x + 2y = 4$ becomes $x - 2y = -4$ under the same multiplication, and both symbols stay strict, so both lines are dashed.

The corner-point comparison has to be run again for each objective, because the corner that wins depends on the weights.

$P: \; 0, \; 25, \; 23, \; 12 \qquad\text{and}\qquad Q: \; 0, \; 5, \; 15, \; 18$

reading the corners in the order given. The heavy weight on $x$ in $P$ sends it to the wide corner $(5, 0)$, worth $25$; the heavy weight on $y$ in $Q$ sends it to the tall corner $(0, 6)$, worth $18$. The region did not change, only the objective, and the same list of corners answers both questions.