1. **Problem Statement:** Find the zeros of the functions algebraically for problems 7 to 12. 2. **Recall:** The zero of a function $f(x)$ is the value of $x$ for which $f(x) = 0$. 3. **Solve each function:** **7.** $f(x) = x + 2$ Set $f(x) = 0$: $$x + 2 = 0$$ Subtract 2 from both sides: $$x + \cancel{2} = \cancel{2} - 2$$ $$x = -2$$ **8.** $f(x) = -2x + 6$ Set $f(x) = 0$: $$-2x + 6 = 0$$ Subtract 6: $$-2x + \cancel{6} = \cancel{6} - 6$$ $$-2x = -6$$ Divide both sides by $-2$: $$\frac{-2x}{\cancel{-2}} = \frac{-6}{\cancel{-2}}$$ $$x = 3$$ **9.** $f(x) = \frac{1}{3}x - 6$ Set $f(x) = 0$: $$\frac{1}{3}x - 6 = 0$$ Add 6: $$\frac{1}{3}x - \cancel{6} = \cancel{6}$$ $$\frac{1}{3}x = 6$$ Multiply both sides by 3: $$3 \times \frac{1}{3}x = 3 \times 6$$ $$x = 18$$ **10.** $f(x) = 2x + 10$ Set $f(x) = 0$: $$2x + 10 = 0$$ Subtract 10: $$2x + \cancel{10} = \cancel{10} - 10$$ $$2x = -10$$ Divide both sides by 2: $$\frac{2x}{\cancel{2}} = \frac{-10}{\cancel{2}}$$ $$x = -5$$ **11.** $f(x) = -\frac{2}{5}x + 4$ Set $f(x) = 0$: $$-\frac{2}{5}x + 4 = 0$$ Subtract 4: $$-\frac{2}{5}x + \cancel{4} = \cancel{4} - 4$$ $$-\frac{2}{5}x = -4$$ Divide both sides by $-\frac{2}{5}$: $$x = \frac{-4}{-\frac{2}{5}} = -4 \times \frac{5}{-2} = 10$$ **12.** $f(x) = 4x$ Set $f(x) = 0$: $$4x = 0$$ Divide both sides by 4: $$\frac{4x}{4} = \frac{0}{4}$$ $$x = 0$$ **Final answers:** 7. $x = -2$ 8. $x = 3$ 9. $x = 18$ 10. $x = -5$ 11. $x = 10$ 12. $x = 0$