Systems of Inequalities

When the answer is a whole region

A6 ended with two lines agreeing at exactly one point. But real constraints rarely say exactly. You’re stocking a study session: snacks cost $$2$, drinks cost $$3$, and you can spend at most $$12$ — that’s $2x + 3y \le 12$. You need at least $4$ items so nobody goes without — that’s $x + y \ge 4$. Can you buy $3$ snacks and $2$ drinks? Run both tests: $2(3) + 3(2) = 12 \le 12$ ✓ (hitting the budget exactly is allowed — at most includes the boundary) and $3 + 2 = 5 \ge 4$ ✓. So $(3, 2)$ works. So does $(4, 1)$. So do plenty of others — and that’s the point: loosen the equals signs and the solution stops being a point. It becomes a region.

One inequality shades half the plane

You’ve seen this move before, one dimension down. In A2, $x \le 5$ wasn’t a dot on the number line — it was a shaded ray, with a circle at the boundary saying whether $5$ itself counted. In two variables, $2x + 3y \le 12$ isn’t a line — it’s a shaded half-plane, and the boundary line (the version with $=$, straight from A4) plays the circle’s role. Watch the whole method on the budget constraint, using only moves you already own:

That’s the whole method: solve for $y$, draw the boundary solid ($\le$, $\ge$) or dashed ($\lt$, $\gt$), shade the side the solved form names, and confirm with a test point. A strict sign gets a dashed boundary for the same reason A2 drew an open circle: the edge itself is the one set of points not invited.

Two tests, one overlap

A system of inequalities runs the same membership test A6 taught — just twice, keeping only the points that pass both. Graph the second constraint the same way: $x + y \ge 4$ isolates to $y \ge -x + 4$ in one move (the $y$-coefficient is already $1$ — there’s nothing to divide). Boundary solid, shade above. And this time the test point tells the opposite story: $(0, 0)$ gives $0 \ge 4$ — false — so the shading must leave the origin out. A failing test point is just as informative as a passing one.

Lay both shadings on one grid and the answer draws itself: the solution region is where the two shadings stack — every point in the doubly-shaded wedge passes both tests, infinitely many of them.

The solver opens on the snack system you just worked. Before reading its steps, predict: which side of each line shades, and does $(0,0)$ pass each test? Then try the chip $2x - y < 4$ — decide which side it shades before looking. If you guessed “below, because it says less than,” watch the divide step carefully. And try $y > 2x + 3$ with $y < 2x - 1$: parallel boundaries whose shadings face away from each other — the two-variable version of “no solution.”

Drive a point through the region

Drag the test point straight down through the wedge and out the bottom: exactly one check flips from ✓ to ✗ as you cross a boundary — a region’s edge is where one test changes its mind. Then press facing away — no solution and hunt for a spot with two ✓ marks: there isn’t one. Now drag the upper line’s $b$ handle down past the other line and watch a band of solutions open up — “no solution” was never about the parallel slopes; it’s about the shadings facing away with a gap. Finally, press a point on the boundary: the point sits exactly on a dashed line, so it fails — now switch that $\lt$ to $\le$ and watch the verdict flip with zero geometry change.

Where the wrong intuitions come from

“Less than means shade below” feels right because it’s usually right — when $y$‘s coefficient is positive. But the side is named by the solved form, and solving $2x - y < 4$ for $y$ means dividing by $-1$: A2’s flip rule fires, the $\lt$ becomes $\gt$, and the shading goes above. When in doubt, the test point never lies: it checks the original inequality, flips and all.

“Parallel boundaries means no solution” is an A6 reflex worth unlearning here. Parallel shadings facing away from each other give no solution — but parallel shadings facing toward each other leave a whole band between the lines. Parallelism alone decides nothing; the shading directions do. And notice what’s gone: no system of inequalities has exactly one solution worth hunting for — any region with any area holds infinitely many points, so the SAT’s question is always “none, or infinitely many?”

Finally, the half-answer trap, inherited straight from A6: being in one shaded half-plane means nothing. A point on the dashed edge of one region, or deep inside just one shading, still fails the system — membership means passing both tests, every time.

The one thing to remember

An inequality in two variables shades half the plane — boundary solid or dashed by the same open/closed logic as A2, side chosen by the solved form (flip on a negative divide!), verified by a test point. A system keeps only the overlap: the region where every test passes. Points are in or out one membership check at a time — and the edge cases live exactly on the boundary lines.

Graphing one inequality: the four moves

A system’s solution set

Run every inequality’s shading on one grid; the solution region is the overlap — the points that pass all the membership tests at once.

Reading a point against a system

To decide if $(x_0, y_0)$ is a solution: substitute it into each inequality and evaluate exactly. Both true → in the region. Any false → out. If it lands exactly on a boundary, the sign type decides: $\le$/$\ge$ counts it, $\lt$/$\gt$ doesn’t.