Solve Each Inequality and Graph Its Solution

To solve each inequality and graph its solution, you run the same four moves every time:

1. Solve it like an equation. Undo operations on both sides — add, subtract, multiply, divide — until the variable stands alone.
2. Watch the sign. If you multiply or divide both sides by a negative number, reverse the inequality symbol.
3. Mark the endpoint. Draw an open circle for $<$ or $>$, and a closed dot for $\leq$ or $\geq$.
4. Shade the direction. Pick one easy number on each side of the endpoint and test it in the original inequality; shade the side that works.

The second move is where most lost points live. Adding or subtracting never changes the symbol: $x + 3 > 5$ becomes $x > 2$, and $x - 7 \leq 1$ becomes $x \leq 8$, no drama either way.

Multiplying or dividing is different. Divide the true statement $10 > 6$ by $2$ and you get $5 > 3$, still true. Divide it by $-2$ and raw arithmetic gives $-5 > -3$, which is false. The only honest repair is to flip it to $-5 < -3$. That is why we solve each inequality and graph its solution with the flip rule in hand: the symbol must keep saying which side is bigger.

One word deserves its own note. Each is a counting word: every listed inequality gets its own full solution and its own graph. A worksheet prompt means committing to the whole list.

A number line graph answers two questions at a glance: where does the solution start, and which way does it run?

The endpoint tells you where. An open circle at $-2$ means $-2$ itself is not a solution — exactly what the strict inequality $x > -2$ demands. A closed dot would mean the endpoint is included, as in $x \leq 3$.

The shading tells you which way. Shading to the right collects every number bigger than the endpoint; shading to the left collects every number smaller. After you solve the inequality and graph, read the picture back as a sentence: "everything to the right of $-2$, endpoint excluded." Ten seconds of testing one shaded value catches almost every direction mistake.

When you solve each inequality and graph its solution, the picture and the algebra must agree on three details at once.

The endpoint style matches the symbol, the shading matches the direction, and nothing else on the line gets drawn.

Practicing how to solve each inequality and graph its solution really means auditing those three details every single time. Graph-first readers can also solve each inequality and graph its solution by testing points on the drawing itself.

Solve and graph: $y - 4 \geq 6$.

• Add 4 to both sides: $y \geq 10$. Adding a positive number never flips the sign.
• Endpoint: a closed dot at 10, because $\geq$ includes 10 itself.
• Direction: shade right, since every number above 10 works. Check one: $11 - 4 = 7 \geq 6$. ✓

Answer: $y \geq 10$ — closed dot at 10, shading to the right.

One step, one flip-check that passed, one graph: that is the whole routine when you solve each inequality and graph its solution at its simplest. Log the pattern now, because Example 2 will stress-test it. Drills this simple are where you rehearse how to solve each inequality and graph its solution before harder ones arrive. One rep like this is enough preparation to solve each inequality and graph its solution with confidence.

Solve and graph: $-6v > 48$.

• Divide both sides by $-6$. Dividing by a negative reverses the symbol: $v < -8$.

Textbooks have warned about exactly this step for over a century. An old analytic geometry text puts it in one line: "changing signs throughout an inequality reverses the inequality sign" (Smith & Gale, 1904).

• Endpoint: an open circle at $-8$, because the strict $<$ excludes it.
• Direction: shade left. Check a shaded value: $-6 \times (-9) = 54 > 48$. ✓

Answer: $v < -8$ — open circle at $-8$, shading to the left.

This example is the classic reason students drop points when they solve each inequality and graph its solution under time pressure. The algebra was one division; the only possible error was the flip, and it is worth a whole letter grade. Whenever you solve each inequality and graph its solution with a negative divisor, say the flip out loud as you write it. The graph makes the flip visible: solve each inequality and graph its solution, and the arrow direction confirms the reversed symbol.

A phone plan costs \24$per month plus$$0.15$per text. Lina can spend at most$$45$ a month. How many texts can she send?

Set up her monthly total and keep it at or under the ceiling:

$24 + 0.15t \leq 45$

• Subtract 24 from both sides: $0.15t \leq 21$.
• Divide by $0.15$ (positive, so no flip): $t \leq 140$.
• Graph: a closed dot at 140 with shading to the left — 140 texts are allowed, and so is every smaller number.

Answer: at most 140 texts per month, i.e. $t \leq 140$.

Budget problems like this one show why it pays to solve each inequality and graph its solution instead of guessing one safe number. The graph displays every acceptable value at once, ceiling included. The same routine lets you solve each inequality and graph its solution for any budget, data plan, or paycheck cap. Money limits are the natural place to solve each inequality and graph its solution before spending a single dollar.

Worksheet instructions are short, so every word carries weight:

• Solve — do the algebra until the variable stands alone.
• Each — every listed item is a separate problem with a separate answer.
• Graph — the answer is not finished until it lives on a number line.
• Its — each solution belongs to its own inequality. Never reuse one graph for the whole list.

Read that way, to solve each inequality and graph its solution is a four-part promise, and graders check all four parts. A sheet headed "solve each inequality and graph its solution" is announcing its rubric in advance. Graders expect you to solve each inequality and graph its solution once per listed item — no more, no less.

Solve and graph: $7 - 3x > 16$.

• Subtract 7 from both sides: $-3x > 9$. Subtraction never flips anything.
• Divide by $-3$ and flip: $x < -3$.
• Endpoint: an open circle at $-3$, because the flipped strict $<$ still excludes it.
• Direction: shade left. Check: $7 - 3(-4) = 19 > 16$. ✓

Two moves instead of one, but the audit is identical: did a negative divisor appear, and did the symbol turn? Ask that every time you solve each inequality and graph its solution, and the flip stops being forgettable.

Answer: $x < -3$ — open circle at $-3$, shading to the left. Any two-step problem asking you to solve each inequality and graph its solution follows this exact script.

Run this list every time you solve each inequality and graph its solution:

1. Algebra check: the endpoint substituted into the original inequality comes out true or exactly equal.
2. Flip check: any step that multiplied or divided by a negative has a reversed symbol.
3. Endpoint check: strict $<$, $>$ get open circles; $\leq$, $\geq$ get closed dots.
4. Direction check: one test value from the shaded side satisfies the original.

Whether you solve each inequality and graph its solution in one step or in four, the list never changes. Make it a reflex and you will solve each inequality and graph its solution correctly on the first pass.

Most lost points on this problem type come from one of four slips:

1. Forgetting the flip. Dividing by a negative without reversing the symbol turns $-6v > 48$ into the wrong answer $v > -8$. The moment your last step divides by a negative, reverse the sign.
2. Drawing the direction backwards. The shading runs toward the values that work, not toward the bigger numbers by default. $v < -8$ shades left, even though left usually means "smaller than everything you solved for."
3. Mixing up open and closed endpoints. A closed dot drawn where an open circle belongs (or the reverse) changes the answer by exactly one, often-graded point: strict $<, >$ take open circles; $\leq, \geq$ take closed dots.
4. Answering only the first item. When the sheet says "each", every listed inequality needs its own solution and its own graph — half the list means half the credit.

Notice that three of the four slips happen after the algebra is already done. Slow, audited work is how strong students solve each inequality and graph its solution without any of these slips.