Solve the equation |-6 + 4x| = |10 + x|.

Step-by-step solution

1. 1

   Tidy each absolute value first. Reordering the terms inside the bars changes nothing, because $|-6+4x| = |4x-6|$ and $|10+x| = |x+10|$. The equation becomes

   $|4x - 6| = |x + 10|$

   This form makes the two linear expressions easy to compare.

2. 2

   Use the rule for two equal absolute values. Squaring both sides gives $A^2 = B^2$, i.e. $(A-B)(A+B) = 0$. So $|A| = |B|$ is equivalent to exactly two linear possibilities:

   $A = B \quad \text{or} \quad A = -B$

   That is why this problem needs two cases and not the four you would get by peeling the bars off independently.

3. 3

   Case 1: the insides are equal.

   $4x - 6 = x + 10$

   $3x = 16 \quad \Longrightarrow \quad x = \frac{16}{3}$

4. 4

   Case 2: the insides are opposites.

   $4x - 6 = -(x + 10) = -x - 10$

   $5x = -4 \quad \Longrightarrow \quad x = -\frac{4}{5}$

5. 5

   Check both candidates in the original equation. Squaring can introduce extraneous roots, so substitution is not optional. For $x = \tfrac{16}{3}$:

   $\left|-6 + \tfrac{64}{3}\right| = \tfrac{46}{3}, \qquad \left|10 + \tfrac{16}{3}\right| = \tfrac{46}{3}$

   For $x = -\tfrac45$:

   $\left|-6 - \tfrac{16}{5}\right| = \tfrac{46}{5}, \qquad \left|10 - \tfrac45\right| = \tfrac{46}{5}$

   Both sides match, so neither root is extraneous.

6. 6

   State the solution set. Two distinct values satisfy the equation:

   $x = \frac{16}{3} \quad \text{or} \quad x = -\frac{4}{5}$

   Geometrically these are the two points where the V-shaped graphs $y = |4x-6|$ and $y = |x+10|$ cross.