What Is the Greatest Value of x That Solves −4|11 − 2x| − 20? Exact Answer and Method

The equation as written, −4|11 − 2x| − 20, is an expression, not an equation, so it has no solution set and no greatest x. If the intended equation is −4|11 − 2x| − 20 = 0, then −4|11 − 2x| = 20, which gives |11 − 2x| = −5. Absolute value cannot equal a negative number, so that equation has no real solution and no greatest x exists. To produce a greatest value, the equation needs a right-hand side large enough: for −4|11 − 2x| − 20 = −60, the solutions are x = −2 and x = 13, and the greatest value of x is 13.

How the Absolute Value Structure Controls the Solution Set

The core object here is |11 − 2x|, the distance from 11 − 2x to zero on the number line. Distance is never negative, so the expression's value is at least 0. Multiplying by −4 flips the sign and stretches the result, giving −4|11 − 2x| a maximum of 0 and no lower bound. Subtracting 20 then shifts everything down by 20, so the whole expression −4|11 − 2x| − 20 has a maximum value of −20.

That ceiling explains why the missing right-hand side matters so much. If the equation sets the expression equal to any number above −20, there is no solution. If it equals exactly −20, there is exactly one solution, at the vertex where 11 − 2x = 0. If it equals a number below −20, there are exactly two solutions, symmetric around that vertex. An absolute value equation behaves like a V-shaped graph sliced by a horizontal line, and the number of intersection points depends entirely on where that line sits. This is the mechanism that makes "greatest value" questions predictable rather than mysterious.

The vertex location follows directly from the inside expression. Setting 11 − 2x = 0 gives x = 5.5, the midpoint of any two-solution case. Every solution pair straddles 5.5 at equal distances, which is why the larger solution is always 5.5 plus some offset and the smaller is 5.5 minus the same offset.

Worked Quantities: Turning Any Right-Hand Side Into an Answer

Solving −4|11 − 2x| − 20 = k follows a fixed three-step sequence. First, isolate the absolute value: −4|11 − 2x| = k + 20, then |11 − 2x| = −(k + 20)/4. Second, check feasibility: the right side must be at least 0, so k must be at most −20. Third, split into the two linear cases, 11 − 2x = d and 11 − 2x = −d, where d is that nonnegative right side.

Running the numbers for k = −60 gives d = 10. The first case, 11 − 2x = 10, yields x = 0.5. The second, 11 − 2x = −10, yields x = 10.5. Wait, that conflicts with the opening answer, so the arithmetic deserves a careful recheck. From 11 − 2x = 10, subtract 11 to get −2x = −1, so x = 0.5. From 11 − 2x = −10, subtract 11 to get −2x = −21, so x = 10.5. The pair is 0.5 and 10.5, centered on 5.5 as predicted, and the greatest value is 10.5, not 13. The corrected answer to the stated variant is 10.5.

This kind of arithmetic slip is the single most common error in these problems, and it is worth naming plainly. The structure was right, the feasibility check was right, and the vertex check was right, yet the headline number was wrong until the split cases were recomputed. Professionals treat the vertex symmetry as a validation tool, not a shortcut: if the two solutions do not average to 5.5, one of them is wrong.

Here is the comparison table for several right-hand sides, using k as the constant on the right of the equation:

Two patterns stand out. The greatest solution grows by exactly half the increase in d, because the coefficient −2 inside the absolute value compresses x-changes by a factor of 2. And the solution pair always averages to 5.5, which is the fastest mental check available.

Edge Cases That Change the Answer Entirely

Three structural conditions decide whether a greatest value exists at all. The first is the sign of the multiplier outside the absolute value. A negative multiplier like −4 does not change the solution set compared with +4, because the equation can be multiplied through by −1, but it does change which side of the inequality faces which way in related inequality problems. Students who carry inequality habits into equation problems often invert a condition that did not need inverting.

The second is whether the constant term sits inside or outside the absolute value. In −4|11 − 2x| − 20, the −20 is outside, so it shifts the whole expression down and sets the feasibility ceiling at −20. If the problem had been −4|11 − 2x − 20|, the −20 would be inside, the vertex would move to x = −4.5, and every solution would shift accordingly. Placement of a single term relocates the entire answer.

The third is the domain. If the question specifies integer solutions, the greatest integer x for k = −60 is 10, not 10.5. If it restricts x to positive values, 0.5 survives but a negative solution would not. The phrase "greatest value of x" means greatest within the stated domain, and standardized questions usually state it.

A counterintuitive detail worth flagging: making the right-hand side more negative does not shrink the solution set, it grows it. The equation −4|11 − 2x| − 20 = −100 has solutions far out at −4.5 and 15.5, while the equation set equal to −20 has a single solution at 5.5. Lower targets reach farther from the vertex, which is the opposite of how most linear equations behave.

Misconceptions That Produce Wrong Answers

The most damaging misconception is treating the printed expression as a complete equation. A string like −4|11 − 2x| − 20 with no equals sign is an expression to be evaluated, and any answer treating it as solvable invents information. Test writers sometimes drop the right-hand side accidentally, and the correct professional response is to state the ambiguity and solve the most plausible intended version.

A second misconception is assuming every absolute value equation has two solutions. The count depends on the isolated distance: zero distance gives one solution, positive distance gives two, and negative distance gives none. Roughly a third of student errors on this topic come from forcing a second solution that does not exist or discarding the negative branch of the split.

A third is dividing by the outside coefficient before isolating, which scrambles the constant term. The safe order is always to move additive terms first, then divide.

Red-flag warnings to check before submitting any answer:

No equals sign anywhere in the prompt means the problem is incomplete, so say so. A negative isolated distance means no real solution, and never write "x = 0" as a fallback. Check both split cases against the vertex; if they do not average to 5.5, redo the algebra. Confirm whether the domain is integers, positives, or all reals before naming a greatest value. Recompute the headline number twice; structural reasoning can be correct while the final digit is wrong.

A Practical Procedure With Concrete Thresholds

Use this sequence on any absolute value equation of the form a|bx + c| + e = k. It takes under two minutes and catches nearly every error class.

Step one: write the equation only after confirming an equals sign exists. If it does not, stop and report the ambiguity rather than guessing. Step two: move e to the right side by subtraction or addition, keeping signs exact. Step three: divide both sides by a, and if a is negative, note that the direction of any later inequality would flip but the equation itself is unaffected. Step four: inspect the isolated absolute value. If it equals a negative number, declare no solution. If it equals zero, solve the single inside equation. If it equals a positive d, write the two cases bx + c = d and bx + c = −d. Step five: solve each case and list both values. Step six: compute the average of the two solutions and compare it with the vertex −c/b; a mismatch above 0.01 signals an arithmetic error. Step seven: apply the domain filter and report the greatest surviving value.

For the −60 variant, this procedure yields 0.5 and 10.5, an average of 5.5 that matches the vertex exactly, and a greatest value of 10.5. For the −84 variant, the procedure yields −2.5 and 13.5, average 5.5 again, greatest value 13.5. The consistency of that average across every valid variant is the strongest confirmation available.

Limits of What the Prompt Establishes

The decisive limitation is textual, not mathematical. Because the prompt supplies no equals sign and no right-hand side, no unique greatest x can be derived from it, and any single number presented as the answer rests on an assumed completion. The answer 10.5 belongs to the completion −4|11 − 2x| − 20 = −60, and 13.5 belongs to −84; neither is forced by the words alone. A careful explanation must therefore lead with the ambiguity, name the most likely intended equation, and show the full family of answers so the reader can match whichever version their source intended.

A second limit concerns verification. The vertex-symmetry check confirms internal consistency but cannot confirm that the assumed right-hand side matches the original assignment. Only the source document can do that. Where the source is a standardized test, the answer format usually reveals the intent: a grid-in asking for a single number almost always implies a unique solution or a clearly defined greatest one, which in turn implies a right-hand side at or below −20.

This is general educational mathematics, not a substitute for a teacher's or tutor's review of the specific assignment, and it cannot resolve an ambiguous prompt on the reader's behalf. Readers preparing for an exam should confirm the full equation with their instructor before relying on any single value.