1. **State the problem:** Find the equation of the line in slope-intercept form $y=mx+b$ that passes through the point $(5,4)$ and is perpendicular to the line given by $2x - 10y = 0$. 2. **Rewrite the given line in slope-intercept form:** $$2x - 10y = 0 \implies -10y = -2x \implies y = \frac{-2x}{-10} = \frac{1}{5}x$$ The slope of the given line is $m = \frac{1}{5}$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal of the original slope. $$m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{1}{5}} = -5$$ 4. **Use point-slope form to find the equation of the perpendicular line:** Point-slope form is: $$y - y_1 = m(x - x_1)$$ Using point $(5,4)$ and slope $-5$: $$y - 4 = -5(x - 5)$$ 5. **Simplify to slope-intercept form:** $$y - 4 = -5x + 25$$ $$y = -5x + 25 + 4$$ $$y = -5x + 29$$ **Final equation:** $$\boxed{y = -5x + 29}$$ --- **Bonus: Solve for $x$ in the equation $14 + 2x = 3|8x - 1|$** 6. **Consider two cases because of the absolute value:** **Case 1: $8x - 1 \geq 0 \implies x \geq \frac{1}{8}$** $$14 + 2x = 3(8x - 1)$$ $$14 + 2x = 24x - 3$$ $$14 + 2x + 3 = 24x$$ $$17 + 2x = 24x$$ $$17 = 22x$$ $$x = \frac{17}{22}$$ Check if $x \geq \frac{1}{8}$: $\frac{17}{22} \approx 0.7727 > 0.125$, valid. **Case 2: $8x - 1 < 0 \implies x < \frac{1}{8}$** $$14 + 2x = -3(8x - 1)$$ $$14 + 2x = -24x + 3$$ $$14 + 2x - 3 = -24x$$ $$11 + 2x = -24x$$ $$11 = -26x$$ $$x = -\frac{11}{26}$$ Check if $x < \frac{1}{8}$: $-\frac{11}{26} \approx -0.423 < 0.125$, valid. **Final solutions:** $$\boxed{x = \frac{17}{22} \text{ or } x = -\frac{11}{26}}$$