Solve |2x−3|=7: Absolute Value Equation

Absolute value equations confuse students because the | | bars hide a fork in the road: the quantity inside can be positive or negative, and both possibilities must be followed. The equation |2x − 3| = 7 does not ask "what makes 2x − 3 equal 7"; it asks "what makes 2x − 3 equal 7 or −7", because both 7 and −7 have absolute value 7.

Once you see the fork, the method is pure routine: split into two ordinary linear equations, solve each, and check both answers in the original. We will find x = 5 and x = −2, verify both, and then look at the elegant "distance" interpretation that lets you solve equations like this in your head.

Worked solution

Step 1: Understand what the bars mean.

|2x − 3| = 7 says: the number (2x − 3) is 7 units away from zero. Two numbers sit 7 units from zero: 7 and −7. So:

either 2x − 3 = 7
or 2x − 3 = −7

This split is the entire method. Everything after it is ordinary linear algebra.

Step 2: Solve the positive case: 2x − 3 = 7.

2x = 7 + 3 = 10
x = 10 ÷ 2 = 5 ✓

First solution: x = 5.

Step 3: Solve the negative case: 2x − 3 = −7.

2x = −7 + 3 = −4
x = (−4) ÷ 2 = −2 ✓

Second solution: x = −2.

Step 4: Check (never skip): test both in the original equation.

For x = 5: |2(5) − 3| = |10 − 3| = |7| = 7 ✓
For x = −2: |2(−2) − 3| = |−4 − 3| = |−7| = 7 ✓

Both satisfy |2x − 3| = 7. The solution set is x = 5 or x = −2.

Step 5: The distance reading (the elegant view).

Divide the original equation by 2 to isolate x inside the bars:

|2x − 3| = 7 → |x − 3/2| = 7/2

Now read it as geometry: |x − 3/2| is the distance from x to 3/2 on the number line. The equation says "x is 7/2 units away from 3/2." Walk 7/2 to the right: 3/2 + 7/2 = 10/2 = 5. Walk 7/2 to the left: 3/2 − 7/2 = −4/2 = −2. Same answers, no algebra. With practice, you can solve |x − a| = b style equations by inspection: the answers are a + b and a − b.

Why this method works

Absolute value measures distance from zero, and every positive distance corresponds to two points: one to the right, one to the left. Splitting |2x − 3| = 7 into 2x − 3 = ±7 is not a trick; it is the definition of absolute value unfolded. The check step matters here because it is easy to write the split down wrong (for example, negating only part of the expression). Substituting back into the original bars catches that instantly, since the bars recompute the distance from scratch.

Common mistakes

Solving only the positive case. Writing x = 5 and stopping is the most common error on this problem type. The bars always hide two cases; if you only find one solution, you are only half done.
Negating incorrectly in the second case. The negative case is 2x − 3 = −7, not −2x + 3 = 7 written carelessly, and not 2x + 3 = −7. Keep the inside of the bars exactly as it is and set the whole thing equal to −7.
"Checking" inside only one case. Plugging x = 5 into 2x − 3 = 7 proves nothing beyond your own linear solving. Check in the original |2x − 3| = 7, where the absolute value does independent work.

FAQ

Q: Can an absolute value equation have just one solution?
A: Yes. |2x − 3| = 0 gives only x = 3/2, because zero is its own negative: the two cases 2x − 3 = 0 and 2x − 3 = −0 are the same equation. And |2x − 3| = −7 has no solution at all, since absolute value can never be negative.

Q: What does |x − 3/2| = 7/2 mean geometrically?
A: It says the distance between x and 3/2 on the number line is 7/2. So x is either 7/2 to the right of 3/2 (which is 5) or 7/2 to the left (which is −2). This picture is worth learning: it turns absolute value equations into one-line mental arithmetic.

Q: How do I handle something like |2x − 3| = |x + 1|?
A: Square both sides (both sides are non-negative, so squaring is safe and reversible): (2x − 3)² = (x + 1)², then solve the quadratic. Or split into the two sign combinations. Either way, check every candidate in the original, because squaring can introduce extraneous roots.

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