1. **State the problem:** We are given two linear equations: $$3x - 4y = 8$$ $$px + 8y + 6 = 0$$ We need to find the value of $p$ such that these two lines are consistent (intersect or are the same line). 2. **Rewrite the second equation:** $$px + 8y = -6$$ 3. **Check for consistency:** For two lines $$a_1x + b_1y = c_1$$ $$a_2x + b_2y = c_2$$ if they are parallel and distinct, then $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ If they intersect, the ratios are not equal. 4. **Calculate ratios:** From the first equation: $a_1=3$, $b_1=-4$, $c_1=8$ From the second equation: $a_2=p$, $b_2=8$, $c_2=-6$ Calculate: $$\frac{a_1}{a_2} = \frac{3}{p}$$ $$\frac{b_1}{b_2} = \frac{-4}{8} = -\frac{1}{2}$$ 5. **Set ratios equal for parallel lines:** $$\frac{3}{p} = -\frac{1}{2}$$ Cross multiply: $$3 \times 2 = -1 \times p$$ $$6 = -p$$ 6. **Solve for $p$:** $$p = -6$$ 7. **Check if lines are distinct:** Calculate $$\frac{c_1}{c_2} = \frac{8}{-6} = -\frac{4}{3}$$ Since $$\frac{a_1}{a_2} = -\frac{1}{2} \neq -\frac{4}{3}$$, the lines are not parallel but intersect. **Final answer:** $$p = -6$$