Naphe the system \( y=5 x+1 \) and \( y<\frac{1}{8} x-4 \)

1. We are given the system \[ y=5x+1 \] and \[ y<\frac{1}{8}x-4. \] 2. Any solution \((x,y)\) must lie on the line \(y=5x+1\) and also satisfy the inequality \[ 5x+1 < \frac{1}{8}x-4. \] 3. To solve the inequality, substitute \(y=5x+1\): \[ 5x+1 < \frac{1}{8}x-4. \] 4. Multiply both sides of the inequality by \(8\) (a positive number, so the inequality direction remains unchanged): \[ 8(5x+1) < 8\left(\frac{1}{8}x-4\right), \] which simplifies to \[ 40x+8 < x - 32. \] 5. Subtract \(x\) from both sides: \[ 40x - x + 8 < -32, \] so \[ 39x + 8 < -32. \] 6. Subtract \(8\) from both sides: \[ 39x < -40. \] 7. Divide by \(39\) (again positive) to isolate \(x\): \[ x < -\frac{40}{39}. \] 8. Thus, the solution set is all points on the line \[ y=5x+1 \] for which \[ x < -\frac{40}{39}. \] Equivalently, the solution set can be written as \[ \{(x,\,5x+1):\; x < -\frac{40}{39}\}. \] 9. For reference, the equality \(5x+1=\frac{1}{8}x-4\) occurs when \[ x=-\frac{40}{39},\quad y=5\left(-\frac{40}{39}\right)+1=-\frac{200}{39}+1=-\frac{161}{39}. \] Since the inequality is strict (\(y<\ldots\)), the point \(\left(-\frac{40}{39}, -\frac{161}{39}\right)\) is not included. The final answer is: \[ \{(x,\,5x+1):\; x < -\frac{40}{39}\}. \]