Overview
An absolute value tells you distance from zero, so its output cannot be negative. That simple meaning supports two different strategies: convert a distance bound into an interval, or split a formula where an expression changes sign. The goal is to choose the strategy that reveals the structure. You will use both to count integer solutions, simplify an expression on a restricted interval, and determine when a parameter produces two positive roots.
Turn a strict distance bound into an open interval
In |2x − 3| < 7, the expression 2x − 3 must lie less than seven units from zero. Write both sides of that information: −7 < 2x − 3 < 7. Adding 3 throughout and dividing by 2 gives −2 < x < 5. Neither endpoint belongs to the solution set because the original comparison is strict.
The question asks for integers rather than the length of the interval. The eligible integers are −1, 0, 1, 2, 3 and 4, so the answer is six. A quick sketch of an open interval can prevent two different errors: counting the boundary values and confusing the distance between boundaries with the number of integers between them. After counting, substitute a boundary into the original expression. Both boundaries give an absolute value of 7, confirming why they are excluded rather than counted.
Let the supplied interval decide each sign
For −3 < a < 0, consider |a| + |a + 3|. The first quantity inside an absolute value is negative; the second is positive. Therefore |a| = −a and |a + 3| = a + 3. Adding gives −a + a + 3 = 3. The variable disappears because every point in this interval has a fixed total distance from the two endpoints.
Do not remove both bars using the same rule merely because the expressions look similar. The sign of a does not determine the sign of a + 3 without considering the shift. The alternative 2a + 3 treats a as nonnegative, while −2a − 3 treats both expressions as negative. Writing the sign of each interior expression before simplifying is a small step that exposes these errors. It also scales to longer expressions whose different absolute values change behavior at different points.
Find the shape before solving for a parameter
Now study 2|x − 1| + |x − 5| = k. Its sign changes occur at x = 1 and x = 5. To the left of 1, the expression is 2(1 − x) + (5 − x) = 7 − 3x. Between 1 and 5 it is 2(x − 1) + (5 − x) = x + 3. To the right of 5 it becomes 3x − 7.
The function decreases toward x = 1 and increases after it, so its minimum value is 4. A horizontal line of height k crosses the graph twice only when k > 4. The requirement that both intersections have positive x adds another restriction. On the left branch the root is (7 − k)/3, which is positive precisely when k < 7. Together these conditions give 4 < k < 7. At k = 4 the two roots merge; at k = 7 the left root is zero and fails strict positivity.
Check boundaries and carry the method forward
The tempting interval 4 < k < 8 recognizes the minimum but places the second cutoff incorrectly. The cutoff comes from the graph’s value at x = 0, because positivity is lost there; it does not come from a sign-change point such as x = 5. This distinction is central to parameter questions: different restrictions can come from different features of the same graph.
When a new problem asks for a number of solutions, resist immediately solving every algebraic branch. First identify turning points, minimum values and the requested domain. Then use equations only where they locate a boundary or intersection. For ordinary absolute-value inequalities, use the same discipline on a smaller scale: identify whether the distance is less than or greater than a threshold, preserve strict endpoints, and check whether the requested answer is a set, a count or a sum. In practice, explain which condition rules out each endpoint before selecting an interval.
Apply the method to a new case
A sensor position x must be within 3 units of 2, but more than 1 unit from −1. Which real positions are allowed? Translate the two distances before calculating: |x − 2| ≤ 3 gives −1 ≤ x ≤ 5, while |x + 1| > 1 gives x < −2 or x > 0. A position must meet both requirements, so intersect the first interval with each branch of the second. The left branch has no overlap; the right branch leaves 0 < x ≤ 5. Keeping the union of the two original conditions would accept positions that satisfy only one requirement. Check the endpoints in the original distances: x = 0 is exactly 1 unit from −1 and is excluded, whereas x = 5 is exactly 3 units from 2 and is included. A point such as x = 2 provides a quick interior check.