Absolute value equations: graph both sides, catch both solutions

An original SAT-style practice question on an absolute value equation, worked step by step in Desmos — the same graphing calculator that's built into Bluebook — with the two-case algebra method alongside.

Practice question

|2x − 7| = x + 4

What is the sum of all solutions to the equation above?

⚡ Desmos is the fastest way here The graph shows every solution at once, so there's no case to forget and no answer to check for being extraneous.

Absolute valueSolving by graphingMedium⏱ ~35 sec

Step by step

Graph each side, typing abs for the bars

Type abs and Desmos turns it into absolute value bars; whatever you type next goes inside them. Put the right-hand side on its own line.

y=abs(2x-7)Shows as y = |2x − 7|, a blue V

y=x+4Red line

What you see: a V with its corner on the x-axis, and a line that cuts through both arms of the V — once on the left, once higher up on the right. If the right-hand crossing is off the screen, zoom out.

Tap both intersections

Tap where the line crosses each arm of the V. Desmos drops a grey point at each intersection and labels its coordinates.

(1,5)Left arm

(11,15)Right arm

What you see: two labelled points. Their x-values, 1 and 11, are the solutions.

Add the solutions

The question asks for the sum of all solutions, so add the two x-values on a new line.

1+11= 12 — done

Answer

The solutions are x = 1 and x = 11, so their sum is 12. Check: |2(1) − 7| = |−5| = 5 and 1 + 4 = 5 ✓; |2(11) − 7| = 15 and 11 + 4 = 15 ✓.

⚠ Common trap

Answering 11 — that's what you get if you drop the absolute value bars and solve only 2x − 7 = x + 4. The negative case gives the second solution, x = 1. Two more slips: answering 3.5, the x-value of the V's corner (where 2x − 7 = 0), which isn't a solution at all; and answering 20 by adding the y-values 5 and 15 instead of the x-values.

The algebra way, if you prefer

Split into two cases: the inside, 2x − 7, equals either x + 4 or −(x + 4).

Case 1: 2x − 7 = x + 4, so x = 11. Case 2: 2x − 7 = −x − 4, so 3x = 3 and x = 1.

Check both: an absolute value can't equal a negative number, so each solution needs x + 4 ≥ 0. Both 11 and 1 pass (15 and 5), so neither is extraneous, and the sum is 12. On versions where one case fails that check, the graph saves you the work: a fake solution simply has no intersection.

More walkthroughs: solving for x by graphing · linear inequalities · systems of linear equations · quadratic roots · vertex and maximum value · circle equations · exponential growth · line of best fit · sum of a quadratic's solutions · Desmos cheat sheet · all walkthroughs

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