Graphing Inequalities: Free Response

Replace the inequality symbol with an equals sign to get the boundary equation, then solve it for $y$.

$2x + y = 4 \;\Longrightarrow\; y = -2x + 4.$

Reading the slope-intercept form off directly gives slope $-2$ and $y$-intercept $(0, 4)$. The original symbol is the strict $>$, so points ON this line are not solutions, which means the boundary is drawn dashed.

Check first that the test point is legal: the boundary $y = -2x + 4$ passes through $(0, 4)$, not through $(0, 0)$, so the origin is a valid choice.

Substitute it into the ORIGINAL inequality, not the boundary equation:

$2(0) + 0 > 4 \;\Longrightarrow\; 0 > 4,$

which is false. A false result means the origin is not a solution, so the entire side it sits on is excluded, and the shaded side is the other one.

The boundary's height at $x = 3$ comes from its equation:

$y = -2(3) + 4 = -2.$

The point $(3, -1)$ has $y = -1$, which is greater than $-2$, so $(3, -1)$ sits above the boundary at that $x$-value. The origin sits below the boundary (its height there is $4$, while the origin's $y$ is only $0$), and part B shaded the side that does NOT contain the origin, the side above the line. Since $(3, -1)$ is on that same above side, it is a solution, and indeed $2(3) + (-1) = 5 > 4$ confirms it directly.

Replacing the symbol with an equals sign gives $x = -1$, a vertical line, since it fixes $x$ with no restriction on $y$. The original symbol $<$ is strict, so draw it dashed.

Test the origin in the original inequality:

$0 < -1,$

which is false, so the origin is excluded, and the shaded region is the other side, everything with $x < -1$, to the left of the line.

The equation $y = 4$ is a horizontal line, since it fixes $y$ with no restriction on $x$. The symbol $\ge$ is inclusive, so the boundary is solid.

Test the origin:

$0 \ge 4,$

which is false, so the shaded side is the one without the origin, everything above the line. The wording at least $4$ means $y$ should be as large as $4$ or larger, which points upward, and the test point confirms exactly that: the two never actually disagree here.

The boundary of $x > 0$ is $x = 0$, which is the $y$-axis itself, and the origin lies exactly on it. Substituting $(0, 0)$ into $x > 0$ gives $0 > 0$, which is false, but that false result comes from landing ON the boundary, not from being cleanly on one side or the other, so it settles nothing about a SIDE.

Choose a different point that is clearly off the line, such as $(1, 0)$:

$1 > 0,$

which is true, so the shaded side is the one containing $(1, 0)$, everything to the right of the line, which also matches the reading of $x > 0$ as greater $x$ lies to the right.

The total raised is $2$ dollars per cookie times the number of cookies, plus $3$ dollars per brownie times the number of brownies, so the total is $2x + 3y$. Wanting at least $90$ dollars means that total should be $90$ dollars or more:

$2x + 3y \ge 90.$

The boundary is the matching equation, $2x + 3y = 90$, and since the symbol is the inclusive $\ge$, the boundary is drawn solid.

Substitute $x = 20$ and $y = 15$ into the total:

$2(20) + 3(15) = 40 + 45 = 85.$

Eighty-five dollars is less than the $90$-dollar goal, by $90 - 85 = 5$ dollars, so the sale does not meet its target, and $(20, 15)$ is not a solution of the club's inequality.

Many combinations work; one is $x = 30$, $y = 10$:

$2(30) + 3(10) = 60 + 30 = 90,$

which meets the goal exactly, since $90 \ge 90$ is true. What makes any pair a solution is nothing about a picture: it is that substituting $x$ and $y$ into $2x + 3y$ and comparing the result with $90$ gives a true statement, precisely the original inequality being tested directly, the same check used in part B.

Check each line on its own.

Line 1 replaces the symbol correctly and reads the solid boundary off the inclusive $\le$: sound. Line 2 substitutes the origin into the first inequality,

$2(0) + 0 \le 6,$

which is true, so it is sound. Line 3 does the same for the second inequality,

$0 > 0 - 1,$

also true, and correctly reads dashed from the strict $>$: sound.

Line 4 is where the reasoning breaks. Both individual regions containing the origin says nothing about combining them by union; a system asks for every one of its inequalities to hold at once, which is the definition of an intersection, not a union. The corrected line reads: the solution of the system is the overlap of the two shaded half-planes, the points that satisfy both inequalities simultaneously. A union would include points satisfying only one inequality, which a system does not allow.

Test each point against BOTH inequalities separately.

For $(1, 1)$:

$2(1) + 1 = 3, \qquad 1 - 1 = 0.$

Since $3 \le 6$ and $1 > 0$, both inequalities hold, so $(1, 1)$ is a solution of the system.

For $(4, -2)$:

$2(4) + (-2) = 6, \qquad 4 - 1 = 3.$

Here $6 \le 6$ holds, but $-2 > 3$ is false, so the second inequality fails. $(4, -2)$ is not a solution of the system, even though it would have been wrongly counted as one under a union reading, since it does satisfy the first inequality alone.

A point belongs to the system's solution only when it satisfies EVERY inequality the system lists, which means it must sit inside EACH individual shaded half-plane at the same time. Writing $R_1$ and $R_2$ for the two shaded half-planes, that membership rule is exactly what an intersection is, and an intersection is always contained in each piece it came from:

$R_1 \cap R_2 \subseteq R_1, \qquad R_1 \cap R_2 \subseteq R_2.$

A union works the opposite way: it keeps a point as soon as it belongs to at least one region, so a union can only be the same size as or larger than either region on its own, $R_1 \subseteq R_1 \cup R_2$.

Line 4 proposes combining the two shaded half-planes by union. Since a union cannot be smaller than either half-plane, and the true solution of a system is an intersection that cannot be bigger than either half-plane, the two descriptions point in opposite directions. That mismatch alone is enough to convict Line 4, with no need to test a single point.

Isolate the $y$-term first:

$4x - 2y \le 8 \;\Longrightarrow\; -2y \le 8 - 4x.$

Dividing both sides by $-2$, a NEGATIVE number, reverses the symbol:

$y \ge \frac{8 - 4x}{-2} = 2x - 4.$

An inequality of the form $y \ge \dots$ keeps every point whose $y$-value is at least as large as the boundary at that $x$, which is the region above the line (together with the line itself).

Isolate the $y$-term:

$4x + 2y \le 8 \;\Longrightarrow\; 2y \le 8 - 4x.$

Dividing both sides by $2$, a POSITIVE number, keeps the symbol as it is:

$y \le \frac{8 - 4x}{2} = 4 - 2x.$

An inequality of the form $y \le \dots$ keeps the region below the line (together with the line itself). The only structural difference from part A is the sign of the coefficient on $y$ before dividing: $-2$ there, $2$ here; every other step is the same.

Start from the given inequality and move the $x$-term to the other side, which never involves multiplying or dividing by anything, so no symbol change happens yet:

$Ax + By \le C \;\Longrightarrow\; By \le C - Ax.$

Now isolate $y$ by dividing both sides by $B$. Since $B \ne 0$, exactly two cases are possible, and no third one exists.

If $B > 0$: dividing an inequality by a POSITIVE number preserves its direction, so

$y \le \frac{C - Ax}{B}.$

The symbol stayed $\le$, and $y \le \dots$ is the region below (or on) the boundary, matching the original $\le$.

If $B < 0$: dividing an inequality by a NEGATIVE number reverses its direction, so

$y \ge \frac{C - Ax}{B}.$

The symbol flipped to $\ge$, which is the region above (or on) the boundary, the opposite of what the original $\le$ would suggest at a glance.

Every real number $B$ that is not zero is either positive or negative, with nothing left over, so these two cases account for every possibility, and the claim is established in full, not merely observed on the two examples of parts A and B.