I’m trying to graph solutions to inequalities on a number line, and my brain keeps flipping the sign at the exact wrong moment (like a pancake mid-air). Here’s the one that got me tangled:
Solve and graph: -2x + 3 ≤ 7 and x – 5 < 2. My attempt: - For -2x + 3 ≤ 7, I did -2x ≤ 4, then divided by -2 and got x ≥ -2 (I flipped the sign because of the negative-pretty sure that’s right?). - For x - 5 < 2, I got x < 7. - Since it’s “and”, I intersected them: I put a closed circle at -2, an open circle at 7, and shaded between them. But then I second-guessed everything: - If the inequality flips, why does the shading go to the right of -2 and not left? My number line starts to look like spilled spaghetti when I think about this. - Do closed vs open circles change if I rewrite the inequality in a different order? Like x ≥ -2 vs -2 ≤ x - are those exactly the same dot style? - When it’s a chained inequality like -3 < 2x + 1 ≤ 7, if I end up dividing by a negative later, do I flip both sides at once or only the one I’m “touching”? - I also tried a similar problem with “or” and accidentally shaded the whole line (oops). Any quick way to tell when I should shade between the points vs outside them? I feel like I’m almost there, but I keep tripping on the direction, the dots (open/closed), and the "and" vs "or" shading. Is there a neat, reliable way to keep these straight? Maybe a sanity check to see if my graph actually matches the inequalities? Any help appreciated!
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3 Responses
You’ve got the first problem exactly right: from −2x + 3 ≤ 7 we get −2x ≤ 4, divide by −2 (flip the sign) to x ≥ −2; from x − 5 < 2 we get x < 7; “and” means intersect, so the solution is all x with −2 ≤ x < 7, i.e., a closed dot at −2, open at 7, shade between. Why shade to the right of −2? Because x ≥ −2 means numbers greater than or equal to −2, which lie to the right; flipping the sign just tells you the final direction (≥ or ≤) after you finish the step. Open vs closed dots: closed for ≤ or ≥, open for < or >, and that never changes if you rewrite the inequality (x ≥ −2 and −2 ≤ x are identical and get the same closed dot). For chained inequalities, treat all three parts the same at each step; if you multiply or divide by a negative, flip both inequality signs at once. Quick example: −3 < 2x + 1 ≤ 7 ⇒ −4 < 2x ≤ 6 ⇒ −2 < x ≤ 3; if it were −3 < −2x + 1 ≤ 7, then −4 < −2x ≤ 6, divide by −2 (negative) to get 2 > x ≥ −3, which is the same as −3 ≤ x < 2. “And” means the overlap (often the segment between the points if the arrows point inward); “or” means everything that satisfies at least one (often the outsides if the arrows point outward). A reliable sanity check: pick a number inside your shaded region (say x = 0 for your first problem) and one outside (say x = 8) and plug them into the original inequalities-inside should make both true for “and,” while outside should fail at least one; endpoints should match open/closed based on <, ≤, >, ≥.
Love this question-nothing like a good inequality to make your brain try a backflip. Let’s tame the pancake-flipping and spaghetti-shading with a few rock-solid rules and a simple routine you can always trust.
First, your specific problem
Solve and graph: -2x + 3 ≤ 7 and x – 5 < 2. -2x + 3 ≤ 7 - Subtract 3: -2x ≤ 4 - Divide by -2 (flip the sign!): x ≥ -2 x – 5 < 2 - Add 5: x < 7 “And” means both must be true at the same time, so we take the intersection: - x ≥ -2 AND x < 7 → the solution is the interval [-2, 7) Graph: - Closed dot at -2 (because ≥ includes -2) - Open dot at 7 (because < excludes 7) - Shade between them Why does the shading go to the right of -2? Because x ≥ -2 means numbers greater than or equal to -2. On a number line, “greater” lives to the right. If you’re ever unsure, plug in a test point: - Try x = -3 (to the left): -2x + 3 = 6 + 3 = 9 ≤ 7? No. So left of -2 is out. - Try x = 0 (to the right): -2(0) + 3 = 3 ≤ 7? Yes. So right of -2 is in. Your graph [-2, 7) is correct. Open vs. closed dots and rewriting inequalities - Open dot: < or >
– Closed dot: ≤ or ≥
– Writing x ≥ -2 is exactly the same as -2 ≤ x. Same dot style, same shading. You’re just reading it from the other side.
Flipping: when and why
– Multiply or divide by a negative → flip the inequality direction.
– Adding or subtracting anything → keep the direction the same.
Intuition: multiplying by -1 reflects the number line across 0, so “bigger” and “smaller” switch roles. That’s why the sign flips.
Chained inequalities: flip both at once if needed
Example 1 (no flip needed):
-3 < 2x + 1 ≤ 7 - Subtract 1 everywhere: -4 < 2x ≤ 6 - Divide by 2 (positive): -2 < x ≤ 3 Example 2 (negative coefficient in the middle; flip both): -3 < -2x + 1 ≤ 7 - Subtract 1 everywhere: -4 < -2x ≤ 6 - Divide by -2 (flip both directions): 2 > x ≥ -3
– If you prefer increasing order, rewrite: -3 ≤ x < 2 The key: whatever operation you do, do it to all parts of the chain, and flip all inequalities if you multiply/divide by a negative. “And” vs. “Or”: when to shade “between” vs. “outside” - AND = intersection = both at once → usually “between” two bounds (like a sandwich), if they form a proper lower and upper bound. - OR = union = at least one → often “outside” if the pieces don’t overlap, but could be a single interval or even the whole line if they overlap. Quick strategy that never fails: 1) Solve each inequality separately. 2) Draw the critical points with the correct dots (open/closed). 3) Shade for each inequality separately. 4) Combine: - AND: keep only where both shadings overlap. - OR: keep any place that has shading from either. Sanity-check with test points This is my favorite low-stress check: - Pick a simple number inside your shaded region and verify it satisfies the original inequalities (not your rearranged ones-go back to the start). - Pick a simple number in each unshaded region and confirm it fails at least one inequality (for AND) or fails both (for OR). - Check the endpoints: if the dot is closed, the endpoint should make the inequality true; if open, it should make it false. A tiny worked example (to practice the rhythm) Solve and graph: 3 − 2x > 7 or x + 4 ≤ 1
1) 3 − 2x > 7
– Subtract 3: -2x > 4
– Divide by -2 (flip): x < -2 2) x + 4 ≤ 1 - Subtract 4: x ≤ -3 Graph each: - x < -2 is open dot at -2, shade left - x ≤ -3 is closed dot at -3, shade left OR = union. The union of “left of -2” and “left of -3” is simply “left of -2.” Final answer: x < -2. Compare with an AND version: 3 − 2x > 7 and x + 4 ≤ 1 gives x < -2 AND x ≤ -3, which is just x ≤ -3 (intersection is the stricter one). Why you accidentally shaded the whole line in an “or” problem If you took x ≥ -2 OR x < 7, that actually covers every real number, since every real number is either ≥ -2 or < 7. Their union is the entire line. A quick way to detect this: if the two individual solution sets overlap and together leave no gaps, you’ll end up shading everything. Summary cheat-sheet - Multiply/divide by negative → flip the sign. Add/subtract → don’t flip. - Open dot: < or >. Closed dot: ≤ or ≥. Rewriting sides doesn’t change dot style.
– AND = overlap (intersection). OR = combine (union).
– For chained inequalities, do the same operation to all parts, and flip all inequality signs if you multiplied/divided by a negative.
– Always sanity-check with a test point inside and outside the shaded region, and test the endpoints.
You’ve got great instincts already-your original graph was spot on. With these little “rituals,” your number line will look like clean geometry, not spaghetti.
You’ve got it: -2x + 3 ≤ 7 ⇒ -2x ≤ 4 ⇒ divide by -2 (flip the sign) ⇒ x ≥ -2, and x − 5 < 2 ⇒ x < 7, so closed dot at −2, open dot at 7, shade between because numbers get bigger to the right; ≤/≥ means closed and means open no matter the order, in a chain flip every inequality if you multiply/divide by a negative (e.g., -3 < 2x + 1 ≤ 7 ⇒ -4 < 2x ≤ 6 ⇒ -2 < x ≤ 3), AND = overlap (between) while OR = union (often the outsides), and sanity-check by testing a point inside (0 works) and outside (10 doesn’t). Quick explainer: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:inequalities/x2f8bb11595b61c86:compound-inequalities/v/compound-inequalities