1. **State the problem:** We are given that $y$ varies directly as $x$, meaning $y = kx$ for some constant $k$. We need to find the unknown values in each case. 2. **Formula and rule:** Direct variation means $y = kx$, where $k$ is the constant of proportionality. To find $k$, use the known values of $x$ and $y$. 3. **Problem 1:** Given $y=12$ when $x=4$, find $y$ when $x=12$. Calculate $k$: $$k = \frac{y}{x} = \frac{12}{4} = 3$$ Use $k$ to find $y$ when $x=12$: $$y = kx = 3 \times 12 = 36$$ 4. **Problem 2:** Given $y=-18$ when $x=9$, find $y$ when $x=7$. Calculate $k$: $$k = \frac{y}{x} = \frac{-18}{9} = -2$$ Find $y$ when $x=7$: $$y = kx = -2 \times 7 = -14$$ 5. **Problem 3:** Given $y=-3$ when $x=-4$, find $x$ when $y=2$. Calculate $k$: $$k = \frac{y}{x} = \frac{-3}{-4} = \frac{3}{4}$$ Use $k$ to find $x$ when $y=2$: $$2 = \frac{3}{4} x \implies x = \frac{2}{\frac{3}{4}} = 2 \times \frac{4}{3} = \frac{8}{3}$$ Intermediate step showing cancellation: $$x = \frac{\cancel{2} \times 4}{\cancel{3}} = \frac{8}{3}$$ 6. **Problem 4:** Given $y=3$ when $x=10$, find $x$ when $y=1.2$. Calculate $k$: $$k = \frac{y}{x} = \frac{3}{10} = 0.3$$ Find $x$ when $y=1.2$: $$1.2 = 0.3 x \implies x = \frac{1.2}{0.3} = 4$$ Intermediate step showing cancellation: $$x = \frac{\cancel{1.2}}{\cancel{0.3}} = 4$$ 7. **Problem 5:** Given $y=2.5$ when $x=0.25$, find $y$ when $x=0.75$. Calculate $k$: $$k = \frac{y}{x} = \frac{2.5}{0.25} = 10$$ Find $y$ when $x=0.75$: $$y = 10 \times 0.75 = 7.5$$ **Final answers:** 1. $y=36$ 2. $y=-14$ 3. $x=\frac{8}{3}$ 4. $x=4$ 5. $y=7.5$