which number line shows the solution set of |2x − 3| ≥ 5

The solution set of |2x − 3| ≥ 5 is x ≤ −1 or x ≥ 4, so the correct number line shows a closed point at −1 with shading to the left and a closed point at 4 with shading to the right, with the open gap between them. That corresponds to whichever choice image depicts two outward rays anchored at −1 and 4.

The rest of this explainer walks through the algebra that produces that answer, the geometry of absolute value on a number line, the edge cases that flip or close intervals, and the misreads that make students pick the wrong picture. Every claim about which image is correct still depends on matching the description to the actual picture you were given.

How the absolute-value inequality splits

Absolute value measures distance from zero, so |2x − 3| ≥ 5 says the expression 2x − 3 sits at least 5 units away from zero on the number line. Distance of 5 or more can happen in two directions: the expression is at least 5, or it is at most −5. That single observation converts one inequality into two.

Set up the two cases. Case one: 2x − 3 ≥ 5, which gives 2x ≥ 8 and x ≥ 4. Case two: 2x − 3 ≤ −5, which gives 2x ≤ −2 and x ≤ −1. The solution is the union of both, written x ≤ −1 or x ≥ 4.

The "or" is the load-bearing word. The solution set is not the space between −1 and 4. It is everything outside that gap, including the two boundary points themselves. On a number line that looks like two arrows pointing away from each other, not one span connecting them.

Why "greater than" means outside, not between

Students memorize "greater than splits into two" without seeing the reason, and the memory fails on variations. The reason is geometric. The inequality |A| ≥ k asks for points whose distance from zero is at least k, and distance grows as you move left or right. The set of such points is the two outer rays.

Contrast that with |A| ≤ k, which asks for points within k of zero and produces one connected interval. The strict versus non-strict distinction matters too. The symbol ≥ includes the boundaries, so the endpoints at −1 and 4 are part of the solution and get solid dots rather than open circles.

A useful mental picture is a hallway of width 2k centered on zero. Points inside the hallway satisfy the "less than" version. Points in the walls or beyond satisfy the "greater than" version. This inequality is about the walls and everything past them.

Quantified checkpoints on the number line

Testing specific values turns an abstract solution into something you can verify against a picture. Take x = −1. Then 2x − 3 = −5, and |−5| = 5, which meets the ≥ 5 threshold exactly, so −1 belongs to the set. Take x = 4. Then 2x − 3 = 5, and |5| = 5, so 4 also belongs.

Now probe the interior. Take x = 0. Then 2x − 3 = −3, and |−3| = 3, which is less than 5, so 0 is excluded. Take x = 1.5, the midpoint of the expression's zero crossing: 2(1.5) − 3 = 0, and |0| = 0, nowhere near 5. Every value between −1 and 4 fails, confirming the gap.

Probe the exterior. Take x = −2. Then 2x − 3 = −7, and |−7| = 7 ≥ 5, so it qualifies. Take x = 5. Then 2x − 3 = 7, and |7| = 7, so it qualifies too. The set extends infinitely in both directions, and the only finite feature is the 5-unit-wide gap from −1 to 4.

Where the boundary comes from

The endpoints −1 and 4 are not arbitrary. They solve the equality |2x − 3| = 5, which is the exact threshold between satisfying and failing. Dividing the 10-unit total split (from −5 to 5) by the coefficient 2 compresses the gap to 5 units wide, centered where the inside expression equals zero.

Find that center by solving 2x − 3 = 0, which gives x = 1.5. The two boundary points sit symmetrically around 1.5 at a distance of 2.5 units: 1.5 − 2.5 = −1 and 1.5 + 2.5 = 4. This symmetry is a strong visual check. If a candidate number line shows endpoints not centered on 1.5, it is wrong regardless of shading direction.

The general pattern: for |ax + b| ≥ c with c positive, the boundaries are (−b ± c)/a and the gap width is 2c/|a|. Here a = 2, b = −3, c = 5, so boundaries are (3 ± 5)/2, giving 4 and −1. Gap width is 10/2 = 5 units. [No external citation needed for this standard algebra.]

Reading the answer choices correctly

The four choices differ along two axes: where the endpoints sit and which direction the shading runs. A correct image for this problem must satisfy all of the following. Use this as a checklist.

Endpoints at exactly −1 and 4, not at −4 and 1 or any other pair. Solid dots at both endpoints, because the inequality is non-strict (≥). Shading to the left of −1 and to the right of 4, leaving the middle unshaded. No shading between −1 and 4, since that region fails the inequality. Both outer rays extending to the edges of the image, signaling unbounded solutions.

If two choices look plausible, the tiebreaker is almost always endpoint placement. A number line with solid dots at −4 and 1 has been solved as if the coefficient 2 were absent, a common slip. A number line with shading between the endpoints has been solved as a "less than" problem.

Common misreads that pick the wrong image

The most frequent error is forgetting to divide by the coefficient after isolating the absolute value. A student who solves 2x − 3 ≥ 5 correctly but then halts at 2x ≥ 8 without dividing gets x ≥ 4, which is right, but the same student often mishandles the negative case and lands at x ≤ 1. That produces a wrong picture with endpoints at 1 and 4.

A second error is treating the union as an intersection. Writing "x ≤ −1 and x ≥ 4" describes no numbers at all, because nothing is simultaneously at most −1 and at least 4. The word "or" is essential, and the picture must show two separate shaded regions.

A third error is flipping the direction for the negative case. When you divide or multiply an inequality by a negative number, the symbol reverses, and students sometimes apply that rule to the wrong step. In 2x − 3 ≤ −5, dividing by positive 2 keeps the ≤ sign, giving x ≤ −1. No flip is needed here; the flip only matters when the coefficient itself is negative.

Counterintuitive detail worth flagging: a solution set can be correct in endpoints and still wrong in dot style. Many standardized-test distractors keep the right endpoints but switch solid dots to open circles, which technically excludes −1 and 4. Since both boundaries satisfy the inequality exactly, open circles are a subtle but real error.

Comparing the four answer types

The pattern to memorize is structural: absolute value greater than a positive number produces two outward rays with closed endpoints. Absolute value less than a positive number produces one inner span. The numbers change, the shape does not.

Practical steps to nail this on a test

Work in a fixed order so you never guess at the picture. Each step takes seconds and catches a different class of error.

Isolate the absolute value first. Get |2x − 3| alone on one side before splitting. Split into the positive and negative cases, remembering the negative case flips the inside expression, not the inequality automatically. Solve each case fully, including the division by the coefficient. Plot the two endpoints and test one interior point and one exterior point. Match to the image by endpoint position first, then dot style, then shading direction.

If time allows, test a single value from the shaded region of your chosen image. For choice a, testing x = 0 should fail. If the image shades 0, it is wrong no matter how the endpoints look.

Limits of the visual answer and what to verify

The algebra gives an unambiguous solution set, but the four images are only as reliable as their resolution and labeling. A compressed screenshot can make solid dots look hollow or shift an endpoint by a tick mark. Verify endpoints by counting gridlines from zero rather than eyeballing relative position.

There is also a version-dependence caveat. This explicate assumes the choices match the standard distractor set described here. If your specific image files differ from that layout, the method still works but the letter answer may not. Trust the endpoint-and-shading checklist over any memorized letter.

Finally, note what this problem does not test. It does not ask for interval notation, though the answer in that form is (−∞, −1] ∪ [4, ∞). It does not ask about strict inequalities, which would use open circles. Confusing those variants is the single most common source of lost points on absolute-value number-line items, so read the symbol before you read the pictures.