Alright I've seen this trip people up a lot. It's one of those things that looks simple but the logic has a few traps if you just try to brute force it or treat it like a standard equality. I actually spent a whole afternoon debugging a data validation script once because of this. The script was supposed to check if a sensor reading was within a certain tolerance `|reading - expected| <= tolerance` and my initial implementation was just totally wrong. It was accepting all sorts of out-of-bounds values. The bug was in how I unwrapped the absolute value. Classic mistake. The key thing you have to burn into your brain is that absolute value is about distance from zero. Once you get that everything else falls into place. `|x|` isn't just "make x positive". It's "how far is x from 0 on the number line". So `|5|` is 5 and `|-5|` is also 5. Both are 5 units away from zero. This distance concept is why you can't just drop the bars and solve. You have to split the problem into cases. There are fundamentally two types of these inequalities and you handle them differently. Case 1: The 'AND' Case (`<` or `<=`) Think about `|x| < 3`. This is asking for all the numbers `x` whose distance from zero is *less than* 3. If you picture a number line these are all the numbers squished between -3 and 3. So `x` must be greater than -3 AND `x` must be less than 3. This gives you the compound inequality: `-3 < x < 3`. The general rule is: If you have `|A| < B` (and B is positive), you rewrite it as `-B < A < B`. Case 2: The 'OR' Case (`>` or `>=`) Now think about `|x| > 3`. This is asking for all the numbers `x` whose distance from zero is *greater than* 3. On a number line these are the numbers that are far out in the tails. So `x` can be greater than 3 (like 4, 5, 6...) OR `x` can be less than -3 (like -4, -5, -6...). This gives you two separate inequalities: `x < -3` OR `x > 3`. The general rule is: If you have `|A| > B` (and B is positive), you rewrite it as `A < -B` OR `A > B`. So here is the step-by-step process I use now. This is the algorithm my data validation script should have used from the start. **My Step-by-Step Solver Logic** 1. **Isolate the Absolute Value Expression.** This is the first thing you always do. Get the `|...|` part by itself on one side of the inequality. Use standard algebra add subtract multiply divide whatever you need. If you have `2|x - 1| + 3 > 9`, you first subtract 3 to get `2|x - 1| > 6`, then divide by 2 to get `|x - 1| > 3`. Now you can proceed. 2. **Check the Constant.** Look at the number on the other side. Let's call it `B`. This is a critical sanity check. * If `B` is negative: * `|A| < B` (e.g. `|x| < -5`): This is impossible. An absolute value is a distance so it can't be negative. The solution is the empty set. No solution. * `|A| > B` (e.g. `|x| > -5`): This is always true. An absolute value is always 0 or positive so it's always greater than any negative number. The solution is all real numbers. * If `B` is zero: * `|A| < 0`: Impossible. No solution. * `|A| <= 0`: Only possible if `A = 0`. Solve for that. * `|A| > 0`: True for everything except where `A = 0`. Solve `A != 0`. * `|A| >= 0`: Always true. The solution is all real numbers. 3. **Apply the Rule.** If `B` is positive then you apply the core logic from above. * For `<` or `<=`, use the 'AND' rule: `-B < A < B`. * For `>` or `>=`, use the 'OR' rule: `A < -B` or `A > B`. 4. **Solve the Resulting Inequality(s).** Now you have one or two standard linear inequalities. Just solve for the variable. Let's try to code this up. Implementing a solver that can parse a string is a bit tricky you have to deal with parsing the expression. Parsing these with regex is possible but as they say if you have a problem and you use regex now you have two problems. For a simple demo we can just create a function that takes the parts of the expression already broken up. Here's a basic python implementation for a simple `|ax + b|` form. ```python # Snippet 1: Basic Solver for a simple case # Assumes the form |ax + b| op C # where op is <, <=, >, >= def solve_simple_abs_inequality(a, b, op, C): """ Solves an inequality of the form |ax + b| op C. 'op' is a string: '<', '<=', '>', '>=' Returns a string describing the solution set. """ print(f"Solving: |{a}x + {b}| {op} {C}") # Step 2: Check the constant C if C < 0: if op in ['<', '<=']: return "No solution (absolute value cannot be less than a negative number)" else: # > or >= return "All real numbers (absolute value is always greater than a negative number)" # Step 3: Apply the rule if op in ['<', '<=']: # This is the 'AND' case: -C <= ax + b <= C # We solve two inequalities: ax + b >= -C AND ax + b <= C lower_bound = (-C - b) / a upper_bound = (C - b) / a # If a is negative, the inequality signs flip, so we swap the bounds if a < 0: lower_bound, upper_bound = upper_bound, lower_bound return f"Solution: {lower_bound} <= x <= {upper_bound}" elif op in ['>', '>=']: # This is the 'OR' case: ax + b <= -C OR ax + b >= C bound1 = (-C - b) / a bound2 = (C - b) / a # If a is negative, the inequality signs flip if a < 0: # ax <= -C becomes x >= (-C-b)/a -> bound1 # ax >= C becomes x <= (C-b)/a -> bound2 return f"Solution: x <= {bound2} or x >= {bound1}" else: return f"Solution: x <= {bound1} or x >= {bound2}" # Example from my explanation # |x| < 3 is |1x + 0| < 3 print(solve_simple_abs_inequality(1, 0, '<', 3)) # |x| > 3 is |1x + 0| > 3 print(solve_simple_abs_inequality(1, 0, '>', 3)) # A more complex one: |2x - 4| >= 6 print(solve_simple_abs_inequality(2, -4, '>=', 6)) ``` That code works for the isolated form. But what about step 1 where you have to isolate it first? That's more of a parsing problem. Let's make another function that can handle a slightly more complex form like `k|ax + b| + d op C`. ```python # Snippet 2: Solver that handles isolation first def solve_abs_inequality_with_isolation(k, a, b, d, op, C): """ Solves an inequality of the form k|ax + b| + d op C. 'k' is the multiplier outside, 'd' is the added term. """ print(f"Solving: {k}|{a}x + {b}| + {d} {op} {C}") # Step 1: Isolate the absolute value # Subtract d from both sides new_C = C - d # Divide by k. This can flip the inequality! new_op = op if k < 0: # Flip the operator if we divide by a negative op_map = {'>': '<', '<': '>', '>=': '<=', '<=': '>='} new_op = op_map[op] # Check for k=0 which is a weird case but good to handle if k == 0: if d > C: return "All real numbers" else: return "No solution" final_C = new_C / k print(f"Isolated form: |{a}x + {b}| {new_op} {final_C}") # Now we can just use our first function return solve_simple_abs_inequality(a, b, new_op, final_C) # Example: 3|x + 2| - 5 > 4 # k=3, a=1, b=2, d=-5, op='>', C=4 # Isolate: 3|x+2| > 9 => |x+2| > 3 print(solve_abs_inequality_with_isolation(3, 1, 2, -5, '>', 4)) # Example with negative multiplier: -2|x - 1| + 7 >= 1 # k=-2, a=1, b=-1, d=7, op='>=', C=1 # Isolate: -2|x-1| >= -6 => |x-1| <= 3 (note the flip!) print(solve_abs_inequality_with_isolation(-2, 1, -1, 7, '>=', 1)) ``` The real beast is when you have absolute values on both sides like `|x - 2| > |2x + 1|`. The case-splitting logic gets messy here. You have to consider regions where each expression is positive or negative. The much better trick is to square both sides. Since an absolute value is always non-negative squaring both sides preserves the inequality. `|A| > |B|` is equivalent to `A^2 > B^2`. This turns the problem into a simple quadratic inequality which is much easier to handle. ```python # Snippet 3: Handling absolute values on both sides by squaring import math def solve_double_abs_inequality(a1, b1, op, a2, b2): """ Solves an inequality of the form |a1*x + b1| op |a2*x + b2| by squaring both sides. This converts it to (a1*x + b1)^2 op (a2*x + b2)^2 which simplifies to a quadratic inequality. """ print(f"Solving: |{a1}x + {b1}| {op} |{a2}x + {b2}|") # (a1*x+b1)^2 = a1^2*x^2 + 2*a1*b1*x + b1^2 # Move everything to one side to get Ax^2 + Bx + C op 0 A = a1**2 - a2**2 B = 2*a1*b1 - 2*a2*b2 C = b1**2 - b2**2 print(f"Quadratic form: {A}x^2 + {B}x + {C} {op} 0") if A == 0: # It's actually a linear inequality if B > 0: bound = -C / B return f"Solution: x {op} {bound}" elif B < 0: op_map = {'>': '<', '<': '>', '>=': '<=', '<=': '>='} new_op = op_map[op] bound = -C / B return f"Solution: x {new_op} {bound}" else: # B is also 0 return "All real numbers" if C > 0 else "No solution" # Find roots of the quadratic Ax^2 + Bx + C = 0 discriminant = B**2 - 4*A*C if discriminant < 0: # Parabola never crosses the x-axis # If A > 0 (opens up), it's always positive # If A < 0 (opens down), it's always negative if A > 0 and op in ['>', '>=']: return "All real numbers" if A < 0 and op in ['<', '<=']: return "All real numbers" return "No solution" root1 = (-B + math.sqrt(discriminant)) / (2*A) root2 = (-B - math.sqrt(discriminant)) / (2*A) # Sort roots r_small, r_big = min(root1, root2), max(root1, root2) # The quadratic is a parabola. The solution is either between the roots or outside the roots. # If A > 0 (parabola opens up), > 0 is outside roots, < 0 is between roots if A > 0: if op in ['>', '>=']: return f"Solution: x <= {r_small} or x >= {r_big}" else: # <, <= return f"Solution: {r_small} <= x <= {r_big}" else: # A < 0 (parabola opens down), > 0 is between roots, < 0 is outside roots if op in ['>', '>=']: return f"Solution: {r_small} <= x <= {r_big}" else: # <, <= return f"Solution: x <= {r_small} or x >= {r_big}" # Example: |x - 2| > |2x + 1| # |1x - 2| > |2x + 1| # (x-2)^2 > (2x+1)^2 => x^2 - 4x + 4 > 4x^2 + 4x + 1 => 0 > 3x^2 + 8x - 3 # We are solving 3x^2 + 8x - 3 < 0 print(solve_double_abs_inequality(1, -2, '>', 2, 1)) ``` This stuff seems purely academic but it shows up. That sensor data problem I had was one example. Another time was in a simple physics simulation for collision detection where I had to check if an object's position was within a certain boundary defined by an absolute value function. Getting the logic wrong meant objects could just pass right through walls. Not great. If you want to go deeper I wouldn't look for blogs. I'd go for solid books. For the pure math foundation of inequalities and absolute values you can't go wrong with something like Michael Spivak's "Calculus". It's rigorous and builds everything from the ground up. For the algorithmic side of things like parsing expressions and formalizing rules something like "Introduction to Algorithms" by Cormen Leiserson Rivest and Stein (CLRS) is the bible. It teaches you to think about problems like this in a structured computer-science way. Hope this helps clear things up. It's really just about breaking it down into those two cases and handling the edge conditions around the constant term. Don't let it psyche you out.

a step-by-step solver for solving inequalities involving absolute values