In short

The function y = |2x - 3| - |x + 1| is piecewise linear — three straight-line pieces glued together at two kinks, one at x = -1 and the other at x = 3/2. Each kink is the breakpoint of one of the absolute value bars. Colour each region differently and the picture becomes a friendly zig-zag instead of a scary symbol-soup. The equation |2x - 3| - |x + 1| = k then has a beautifully visual reading: where does the zig-zag cross the horizontal line y = k? Depending on k, the answer can be zero, one, two, or three points.

You have already met absolute value as distance from zero in Absolute Value — Equations. One pair of bars is fine — you split into two cases and you are done. But what if there are two sets of bars in the same expression, like |2x - 3| - |x + 1|? Now you have two breakpoints, three regions, and a graph that bends twice.

This satellite article is about seeing that graph. Once you can sketch y = |2x - 3| - |x + 1|, every equation of the form |2x - 3| - |x + 1| = k becomes a one-second visual question: how many times does my horizontal line y = k cross the zig-zag?

This skill is exactly what CBSE Class 11 (Functions chapter) and JEE Mains keep coming back to — recognising that a sum or difference of absolute values is just a piecewise linear function in disguise.

Find the breakpoints

Each absolute value bar hides a fork in the road. The fork only matters at the exact value of x where the inside of the bars changes sign — the breakpoint.

Why: an absolute value |u| equals u when u \ge 0 and equals -u when u < 0. The "rule" that defines |u| literally changes at u = 0. That change of rule is what produces the kink in the graph.

Two breakpoints divide the real line into three regions:

  1. x < -1 — both insides negative.
  2. -1 \le x < 3/2 — only x + 1 is non-negative, 2x - 3 is still negative.
  3. x \ge 3/2 — both insides non-negative.

In each region, neither bar will flip again. That means inside each region, y = |2x - 3| - |x + 1| is just a normal linear expression — no bars left, no surprises.

Two breakpoints split the number line into three regionsA horizontal number line. Two tick marks at minus one and three over two divide the line into three coloured zones: red on the left, green in the middle, blue on the right. x = −1 x = 3/2 x < −1 −1 ≤ x < 3/2 x ≥ 3/2
The breakpoints $-1$ and $3/2$ split the number line into three regions. Each region gets its own colour and its own simple linear formula.

Compute the formula in each region

Now you remove the bars, region by region. The rule is simple: if the inside is non-negative, the bars do nothing; if the inside is negative, the bars flip the sign.

Worked example 1 — the three formulas

Region 1: x < -1. Both insides are negative.

  • 2x - 3 < 0, so |2x - 3| = -(2x - 3) = -2x + 3.
  • x + 1 < 0, so |x + 1| = -(x + 1) = -x - 1.
y = (-2x + 3) - (-x - 1) = -2x + 3 + x + 1 = -x + 4.

Region 2: -1 \le x < 3/2. Now x + 1 \ge 0 but 2x - 3 is still negative.

  • |2x - 3| = -(2x - 3) = -2x + 3.
  • |x + 1| = x + 1.
y = (-2x + 3) - (x + 1) = -2x + 3 - x - 1 = -3x + 2.

Region 3: x \ge 3/2. Both insides are non-negative; the bars do nothing.

  • |2x - 3| = 2x - 3.
  • |x + 1| = x + 1.
y = (2x - 3) - (x + 1) = 2x - 3 - x - 1 = x - 4.

So the full piecewise picture is

y = \begin{cases} -x + 4, & x < -1 \\ -3x + 2, & -1 \le x < 3/2 \\ x - 4, & x \ge 3/2. \end{cases}

Sanity check at the kinks: at x = -1, region 1 gives -(-1)+4 = 5 and region 2 gives -3(-1)+2 = 5 — they match. At x = 3/2, region 2 gives -3(3/2)+2 = -5/2 and region 3 gives 3/2 - 4 = -5/2 — they match. Why: the function |2x-3| - |x+1| is continuous everywhere, so its piecewise pieces must agree at the breakpoints. If your formulas disagree at a kink, you made an arithmetic slip — go back and check.

Notice the slopes: -1, then -3, then +1. Each time you cross a breakpoint, the slope changes — that is the kink. The slope change at x = -1 is -3 - (-1) = -2, exactly the coefficient of x inside |x+1| doubled (because flipping a + sign on x to a - sign on x inside the bars subtracts 2x worth of slope). At x = 3/2, the slope jumps from -3 to +1, a change of +4, again twice the coefficient of x inside |2x-3|.

Plot the piecewise

Now glue the three line segments together and add a horizontal line y = k on top. The picture tells you everything.

Colour-coded graph of y equals absolute value of two x minus three minus absolute value of x plus one, with a horizontal line y equals kA piecewise linear graph in three colours. A red segment with slope minus one for x less than minus one, a green segment with slope minus three between minus one and three over two, and a blue segment with slope plus one for x greater than three over two. The minimum value is minus five over two at x equals three over two. A dashed horizontal line at y equals one is shown crossing the graph at two points. x y 6 4 2 −2 −6 −3 −1 3/2 4 (−1, 5) (3/2, −5/2) y = k = 1 y = −x + 4 y = −3x + 2 y = x − 4
The graph of $y = |2x - 3| - |x + 1|$ in three colours. Red for $x < -1$ (slope $-1$), green for $-1 \le x < 3/2$ (slope $-3$), blue for $x \ge 3/2$ (slope $+1$). The minimum value $-5/2$ happens at the right-hand kink. The dashed purple line $y = 1$ cuts the graph at two points.

A few things to notice from the picture:

So the equation |2x - 3| - |x + 1| = k has:

Worked example 2 — count intersections for $k = 1$

Draw the horizontal line y = 1 (the dashed purple line in the figure). Where does it meet the zig-zag?

  • Red piece (x < -1, y = -x + 4): set -x + 4 = 1 \Rightarrow x = 3. But 3 is not in the region x < -1, so reject.
  • Green piece (-1 \le x < 3/2, y = -3x + 2): set -3x + 2 = 1 \Rightarrow x = 1/3. Check: -1 \le 1/3 < 3/2 — yes. Accept.
  • Blue piece (x \ge 3/2, y = x - 4): set x - 4 = 1 \Rightarrow x = 5. Check: 5 \ge 3/2 — yes. Accept.

Two solutions: x = 1/3 and x = 5. The graph confirms: the dashed line meets only the green and the blue, exactly as the algebra predicts.

Why: every time you solve a piecewise equation, you must check that each candidate x actually lives in the region whose formula you used. A candidate from the red formula that does not satisfy x < -1 is a ghost — it would only be valid if the rule extended, but the rule changed.

Worked example 3 — solve for $k = -3$

Now try the line y = -3. The minimum of the function is -5/2 = -2.5, and -3 < -2.5. So the horizontal line sits below the lowest point of the graph and cannot cross it at all. You should expect no solutions. Let's verify by chasing each region.

  • Red (x < -1): -x + 4 = -3 \Rightarrow x = 7. Reject (7 \not< -1).
  • Green (-1 \le x < 3/2): -3x + 2 = -3 \Rightarrow x = 5/3. Reject (5/3 \not< 3/2, since 5/3 \approx 1.67).
  • Blue (x \ge 3/2): x - 4 = -3 \Rightarrow x = 1. Reject (1 \not\ge 3/2).

Every candidate fails the region test. The equation |2x - 3| - |x + 1| = -3 has no real solutions, exactly as the picture told you in advance.

Whenever a JEE Mains problem asks for the number of real solutions of an equation involving two or three absolute values, your fastest move is the same: find the breakpoints, compute each piece, sketch the zig-zag, and count crossings with the line y = k. The algebra is a check, not a search — the picture has already given you the answer.

References

  1. NCERT, Mathematics Class XI, Chapter 2: Relations and Functions.
  2. Wikipedia, Piecewise linear function.
  3. Wikipedia, Absolute value.
  4. Khan Academy, Graphs of absolute value functions.
  5. Paul's Online Math Notes, Absolute Value Equations.