Naphe the system $$ y=5 x+1 $$ and $$ y

Question

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

UpStudy AI · Accepted Answer

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:  
$$
\{(x,\,5x+1):\; x < -\frac{40}{39}\}.
$$
All points on the line $$ y = 5x + 1 $$ where $$ x < -\frac{40}{39} $$.

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

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