The expression as written, −4|11 − 2x| − 20, is not actually an equation, so it has no solution and no greatest x. If the intended equation is −4|11 − 2x| − 20 = 0, the greatest solution is x = 8. A common corrected form, −4|11 − 2x| = −20, gives solutions x = 3 and x = 8, so the greatest value is x = 8. The entire answer hinges on which right-hand side the problem intended.
How absolute value equations actually work
An absolute value measures distance from zero on the number line, so |k| equals k when k is zero or positive and equals −k when k is negative. That definition is why an equation like |u| = c has two cases when c > 0: u = c and u = −c. When c = 0 there is exactly one case, u = 0. When c < 0 there are no real solutions, because no distance is negative.
The structure of this problem adds a wrinkle. The absolute value is multiplied by −4 before anything else happens. Multiplication by a negative number flips the sign of the whole expression, so the left side −4|11 − 2x| is always zero or negative, never positive. That single fact controls the entire solution set, and it explains why the right-hand side matters so much.
Isolating the absolute value is the standard first move. Starting from −4|11 − 2x| = −20, divide both sides by −4 to get |11 − 2x| = 5. Dividing by a negative reverses the direction of an inequality but leaves an equation unchanged, which is a frequent point of confusion. From |11 − 2x| = 5, the two cases are 11 − 2x = 5 and 11 − 2x = −5.
Solving the first case: 11 − 2x = 5 gives −2x = −6, so x = 3. Solving the second: 11 − 2x = −5 gives −2x = −16, so x = 8. Checking both in the original, at x = 3 the inside is 11 − 6 = 5, absolute value 5, times −4 is −20, matching the right side. At x = 8 the inside is 11 − 16 = −5, absolute value 5, times −4 is −20, also matching. Both solutions are genuine, and 8 is the larger.
The reason two answers appear is geometric. The expression 11 − 2x hits the value 5 and the value −5 at two different x-values because it is a straight line with slope −2. A line crosses any horizontal level exactly once, so the two absolute value cases produce one crossing each. That is the mechanism behind the pair of solutions, and it generalizes to every linear absolute value equation with a positive constant.
Why the written problem is ambiguous
The phrase as given contains no equals sign, which means it is an expression, not a sentence that can be true or false. An expression can be simplified or evaluated, but it cannot have solutions. This is not a nitpick. The greatest value of x only exists relative to some equation, so the missing right-hand side is the whole problem.
Three natural readings exist, and each produces a different answer. This table lays them out side by side.
| Intended equation | Isolated form | Solutions for x | Greatest x |
|---|---|---|---|
| −4|11 − 2x| − 20 = 0 | |11 − 2x| = −5 | none (real) | no solution |
| −4|11 − 2x| = −20 | |11 − 2x| = 5 | 3 and 8 | 8 |
| −4|11 − 2x| − 20 = −20 | |11 − 2x| = 0 | 5.5 | 5.5 |
In the first row, isolating gives |11 − 2x| = −5, which is impossible because absolute value is never negative. In the second, the constant is positive after isolation and two solutions emerge. In the third, the constant collapses to zero and the single solution comes from 11 − 2x = 0.
Textbook and online homework systems most often intend the second reading, where the equation is set equal to −20 and the standalone −20 at the end of the prompt is actually the right-hand side. That convention is why x = 8 is the answer most answer keys reward. If your source shows a different equals sign, substitute it and re-solve with the same three-step method.
Edge cases that change the count of solutions
The number of solutions depends entirely on the constant that survives isolation. Call that constant c, so the isolated form is |11 − 2x| = c. When c > 0, there are two solutions. When c = 0, there is one. When c < 0, there are none in the real numbers. Memorizing those three cases is faster than re-deriving them each time.
A subtle edge case involves the coefficient on x. Here the inside is 11 − 2x, and the −2 means the expression decreases as x increases. That is why the case 11 − 2x = 5 yields the smaller x and the case 11 − 2x = −5 yields the larger x. Students who assume the positive case always gives the bigger answer get this backward. Verify by substituting rather than by assumption.
Another edge case is a variable on the right side, as in −4|11 − 2x| = x. Then the sign of the right side constrains everything, because the left side is never positive, so any solution must satisfy x ≤ 0. You must also check each candidate in the original equation, since squaring or case-splitting can introduce values that fail. For the clean constant-right-side version, checking is quick and worth doing every time.
One counterintuitive detail deserves emphasis. The expression −4|11 − 2x| − 20 reaches its maximum value of −20 exactly when the absolute value is zero, at x = 5.5. So the whole left side lives at or below −20. That means an equation set equal to any number above −20, such as −4|11 − 2x| − 20 = 10, has no solution at all, while an equation set equal to −20 has exactly one. The ceiling on the expression silently governs solvability.
Misconceptions worth correcting
The most common error is dropping the absolute value bars and solving only 11 − 2x = 5. That produces x = 3 and misses x = 8 entirely, and it is the single biggest source of lost points on this problem type. Always split into both cases before solving.
A second misconception treats the −4 as irrelevant to the sign analysis. Because the multiplier is negative, the left side can never be positive, so equations like −4|11 − 2x| = 20 are impossible before any algebra begins. Spotting that instantly saves time and prevents fabricated solutions.
A third error is distributing the −4 inside the absolute value, writing −44 + 8x. That changes the meaning of the expression and produces wrong answers. The factor outside the bars must stay outside until the bars are isolated.
Red-flag warnings to watch for: No equals sign in the prompt means there is nothing to solve. A negative constant after isolation means no real solution, not a sign error to force. Dividing by a negative flips inequalities but not equations, so do not flip the equals sign. Skipping the substitution check lets extraneous solutions survive in variable-right-side problems. Assuming the positive case always yields the larger x is false when the x-coefficient is negative.
A step-by-step procedure with thresholds
Treat every absolute value equation the same way, and the ambiguity in the written prompt stops mattering. The method below handles all three readings in the table and any variant a textbook throws at you.
Isolate the absolute value first, moving constants and dividing out coefficients in the standard order of operations reversed. Then identify the surviving constant c and classify it as positive, zero, or negative. Positive gives two cases, zero gives one, and negative gives none. Solve each case, then substitute every candidate back into the original equation and discard failures.
For this problem, the practical thresholds are concrete. If c = 5, the solutions are exactly x = 3 and x = 8, and the greatest is 8. If c = 0, the solution is x = 5.5. If c is any negative number, the solution set is empty. If the right side were a variable, restrict candidates by the sign of that right side before checking.
A quick verification habit closes the loop. Compute the inside value at each candidate, take its absolute value, then apply the outside multiplier and constant. For x = 8: inside is −5, absolute value 5, times −4 is −20, which must equal the right side. That 30-second check catches nearly every arithmetic slip.
Reading the answer the way a grader will
Answer keys usually expect the greatest solution when they ask for it, which is why x = 8 is the target in the standard reading. If your equation has two solutions and the question asks for the greatest, you report the larger number only, not both. If it asks for the sum or product, combine them, so the sum here would be 11 and the product 24.
Pay attention to whether the problem wants exact or approximate values. All the answers here are exact integers or a clean decimal (5.5), so no rounding is needed. If a problem produces a fraction, leave it as a fraction unless told otherwise, since decimal rounding can lose a point.
The lesson generalizes beyond this one prompt. The greatest value of x in an absolute value equation is determined by three things: the isolated constant, the sign of the outside multiplier, and the direction of the linear expression inside the bars. Understand those three and you can solve any version of this problem in under a minute.
The article above is general educational information about solving absolute value equations, not a diagnosis, legal opinion, or investment recommendation. It does not promise any particular grade or outcome. For coursework-specific rules, consult your instructor or the textbook's answer key.