1. **Problem 1: Check if ordered pairs satisfy the equation $3x - y = 5$.** The equation is $3x - y = 5$. We substitute each ordered pair $(x,y)$ into the equation and check if the equality holds. **a.** For $(2,1)$: $$3(2) - 1 = 6 - 1 = 5$$ Since $5 = 5$, the pair $(2,1)$ satisfies the equation. **b.** For $(1,4)$: $$3(1) - 4 = 3 - 4 = -1$$ Since $-1 \neq 5$, the pair $(1,4)$ does not satisfy the equation. **c.** For $(0,-5)$: $$3(0) - (-5) = 0 + 5 = 5$$ Since $5 = 5$, the pair $(0,-5)$ satisfies the equation. 2. **Problem 2: Find three solutions of the linear equation $2x + y = 10$.** We will find $y$ for given $x$ values and $x$ for a given $y$ value. The equation is: $$2x + y = 10$$ **Step 1:** Let $x = 1$, solve for $y$: $$2(1) + y = 10$$ $$2 + y = 10$$ $$y = 10 - 2$$ $$y = 8$$ So, one solution is $(1,8)$. **Step 2:** Let $x = 4$, solve for $y$: $$2(4) + y = 10$$ $$8 + y = 10$$ $$y = 10 - 8$$ $$y = 2$$ So, another solution is $(4,2)$. **Step 3:** Let $y = -2$, solve for $x$: $$2x + (-2) = 10$$ $$2x - 2 = 10$$ $$2x = 10 + 2$$ $$2x = 12$$ $$x = \frac{12}{2}$$ $$x = 6$$ So, the third solution is $(6,-2)$. **Final answers:** - Problem 1: $(2,1)$ and $(0,-5)$ satisfy the equation; $(1,4)$ does not. - Problem 2: Three solutions are $(1,8)$, $(4,2)$, and $(6,-2)$.