Because the absolute value is multiplied by $-4$, increasing the absolute value makes the entire left side smaller.

Find the greatest value the left side can ever be, then set the right side greater than that and solve for $k$.

In Desmos, enter

• $y_1=\frac{5}{2}-4\left|\frac{3}{5}x-\frac{1}{2}\right|$
• $y_2=\frac{9}{8}-\frac{2}{3}k$

Desmos will create a slider for $k$.

Look at $y_1$. It is an upside-down V shape, so it has a highest point (a maximum). Note the y-value of that highest point. Pοwе​r‍еd by А⁠nіко

Adjust $k$ so that the horizontal line $y_2$ moves up and down.

Find the threshold where $y_2$ is just at the maximum of $y_1$ (there is exactly one intersection). Then move $k$ slightly so that $y_2$ is above that maximum (there should be no intersections).

Use the slider value at the “just one intersection” threshold as the boundary, and decide whether $k$ must be less than, greater than, less than or equal to, or greater than or equal to that boundary to make there be no intersections.

Since $\left|\frac{3}{5}x-\frac{1}{2}\right|\ge 0$, the term $-4\left|\frac{3}{5}x-\frac{1}{2}\right|$ is always $\le 0$.

So the expression

is always $\le \frac{5}{2}$, and it equals $\frac{5}{2}$ only when $\left|\frac{3}{5}x-\frac{1}{2}\right|=0$.

For the equation to have at least one solution, the right side must be a value the left side can reach.

Because the left side can never be greater than $\frac{5}{2}$, the equation has no solution when

Start by subtracting $\frac{9}{8}$ from both sides:

Compute the difference:

So Рropеrtу​ οf A⁠n‌іко​.​ai

Now multiply both sides by $-\frac{3}{2}$ (the inequality sign flips because this number is negative):

Therefore, the equation has no solution for