Rewrite $|3z+1| = 7$ as two separate equations without the absolute value bars. One equation sets $3z+1$ equal to 7, and the other sets $3z+1$ equal to the opposite of 7.

Once you have the two linear equations, solve each one step by step: first subtract 1 from both sides, then divide by 3. Each equation will give you a possible value of $z$.

After you find the possible values of $z$, plug them back into $|3z+1|$ to verify that the result is indeed 7.

In Desmos, type y = |3x + 1| on one line and y = 7 on another line so you can see where the graphs intersect.

Look for the points where the V-shaped graph of $y = |3x+1|$ crosses the horizontal line $y = 7$. Tap or click on each intersection; Desmos will show the coordinates $(x,y)$ of those points.

The $x$-coordinates of the intersection points are the solutions to $|3z+1| = 7$. Either of those $x$-values is a valid value of $z$ for the equation.

The equation is

Absolute value means distance from 0. So $|3z+1| = 7$ says that the expression $3z+1$ is 7 units away from 0 on the number line. That can happen in two ways: $3z+1$ can be 7 or $-7$.

From $|3z+1| = 7$, write the two possible cases:

• Case 1: $3z + 1 = 7$
• Case 2: $3z + 1 = -7$

Each case is now a simple linear equation in $z$.

Solve each case separately:

For Case 1:

For Case 2:

Now divide each equation by 3 to isolate $z$.

Divide both sides of each equation by 3:

• From $3z = 6$, we get $z = 6/3 = 2$.
• From $3z = -8$, we get $z = -8/3$.

Both $z = 2$ and $z = -8/3$ make $|3z+1| = 7$ true. The question asks for one possible value of $z$, so you can give $\boxed{2}$ as a correct answer.