Solving |2x − 4| = x + 1 Graphically and Algebraically

This question links algebra to a picture: the solutions of $|2x - 4| = x + 1$ are where the V-shaped graph $y = |2x - 4|$ meets the line $y = x + 1$. Solve the two cases, then let the graph confirm how many solutions are genuine.

Solve $|2x - 4| = x + 1$. [5]

The working

Step 1, split into two cases: $2x - 4 = x + 1 \qquad \text{or} \qquad 2x - 4 = -(x + 1) \quad \text{(M1)}$

Step 2, solve each: $2x - 4 = x + 1 \;\Rightarrow\; x = 5 \quad \text{(A1)}$ $2x - 4 = -x - 1 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1 \quad \text{(A1)}$

Step 3, check both (the line $y = x + 1$ must be $\ge 0$ at a solution):

• $x = 5$: LHS $= |10 - 4| = 6$, RHS $= 6$. ✓
• $x = 1$: LHS $= |2 - 4| = 2$, RHS $= 2$. ✓

Both are valid, matching the graph, where the line crosses the V twice: $x = 1 \quad \text{and} \quad x = 5 \quad \text{(M1, A1)}$

Where the marks are won and lost

• The graphical picture tells you to expect two intersections, so two solutions are anticipated. If the algebra gave only one, that would flag an error.
• Both candidates pass the check because $x + 1 \ge 0$ at each ($x = 1$ and $x = 5$ are both $\ge -1$). Had a candidate made the RHS negative, it would be rejected.
• The second case is $2x - 4 = -(x + 1)$, distribute the minus sign to both terms: $-x - 1$.

Common mistakes

• Distributing the negative to only one term: $-(x+1) = -x + 1$ (wrong).
• Skipping the check when it is needed (here both pass, but the habit protects marks elsewhere).
• Giving one solution when the graph clearly shows two crossings.

Full method: Solving Equations Graphically notes. See also Graphs of Modulus Functions. Topic home: Equations, Inequalities and Graphs pillar.