Solve |2a + 7| + |2a - 1| = 8.

Step-by-step solution

1. 1

   Locate the kink points. Each absolute value switches formula where its inside vanishes:

   $2a+7=0\Rightarrow a=-\frac72,\qquad 2a-1=0\Rightarrow a=\frac12$

   These split the line into $a<-\tfrac72$, $-\tfrac72\le a<\tfrac12$, and $a\ge\tfrac12$.

2. 2

   Predict the shape of the answer before solving. Note that $|2a+7|+|2a-1|=2\left(\left|a+\tfrac72\right|+\left|a-\tfrac12\right|\right)$, i.e. twice the total distance from $a$ to $-\tfrac72$ and to $\tfrac12$. The distance between those two points is $4$, so the minimum possible value of the left side is $2\times4=8$ — exactly the right-hand side. That means the equation asks for the minimum, which is attained on a whole segment, not at isolated points.

3. 3

   Case 1: a < -7/2 (both insides negative). Then

   $(-2a-7)+(-2a+1)=-4a-6=8\Rightarrow-4a=14\Rightarrow a=-\frac72$

   This contradicts the strict condition $a<-\tfrac72$, so this case yields nothing new (the value $-\tfrac72$ is picked up in case 2).

4. 4

   Case 2: -7/2 <= a < 1/2 (mixed signs). Here $|2a+7|=2a+7$ and $|2a-1|=-2a+1$, so the $a$ terms cancel:

   $(2a+7)+(-2a+1)=8$

   This reduces to the identity $8=8$, true for every $a$ in the interval. So the entire interval $\left[-\tfrac72,\tfrac12\right)$ is a solution set.

5. 5

   Case 3: a >= 1/2 (both insides non-negative).

   $(2a+7)+(2a-1)=4a+6=8\Rightarrow a=\frac12$

   This satisfies $a\ge\tfrac12$, so $a=\tfrac12$ is included.

6. 6

   Union the cases and state the answer. Case 2 gives $\left[-\tfrac72,\tfrac12\right)$ and case 3 adds the endpoint $\tfrac12$:

   $-\frac72\le a\le\frac12$

   Outside this closed interval the left side exceeds $8$, so nothing else qualifies.

7. 7

   Verify by scanning. Evaluating the left side at $12001$ points from $-6$ to $6$: it equals $8$ at exactly the points of $\left[-\tfrac72,\tfrac12\right]$ and is strictly larger everywhere else ✓.