1. The problem asks to determine if the ordered pair (2, -2) is a solution to the system: $$\begin{cases} 3x + y = 4 \\ x - 3y = -4 \end{cases}$$ 2. Substitute $x=2$ and $y=-2$ into each equation: $$3(2) + (-2) = 6 - 2 = 4$$ $$2 - 3(-2) = 2 + 6 = 8 \neq -4$$ 3. Since the second equation is not satisfied, (2, -2) is not a solution. --- 2. Check if (3, -1) solves: $$\begin{cases} x - 2y = 5 \\ 2x - y = 7 \end{cases}$$ Substitute $x=3$, $y=-1$: $$3 - 2(-1) = 3 + 2 = 5$$ $$2(3) - (-1) = 6 + 1 = 7$$ Both true, so (3, -1) is a solution. --- 3. Check if (-1, 5) solves: $$\begin{cases} -x + y = 6 \\ 2x + 3y = 13 \end{cases}$$ Substitute $x=-1$, $y=5$: $$-(-1) + 5 = 1 + 5 = 6$$ $$2(-1) + 3(5) = -2 + 15 = 13$$ Both true, so (-1, 5) is a solution. --- 4. Solve by graphing: $$\begin{cases} y = \frac{1}{2}x \\ y = -x + 3 \end{cases}$$ Equation 1 slope $m=\frac{1}{2}$, y-intercept $b=0$. Equation 2 slope $m=-1$, y-intercept $b=3$. Set equal to find intersection: $$\frac{1}{2}x = -x + 3$$ $$\frac{1}{2}x + x = 3$$ $$\frac{3}{2}x = 3$$ $$x = \frac{3}{\frac{3}{2}} = 2$$ Substitute $x=2$ into $y=\frac{1}{2}x$: $$y = \frac{1}{2} \times 2 = 1$$ Solution is $(2,1)$. --- 5. Solve by graphing: $$\begin{cases} y = x - 2 \\ 2x + y = 1 \end{cases}$$ Equation 1 slope $m=1$, y-intercept $b=-2$. Rewrite Equation 2: $$y = 1 - 2x$$ Slope $m=-2$, y-intercept $b=1$. Set equal: $$x - 2 = 1 - 2x$$ $$x + 2x = 1 + 2$$ $$3x = 3$$ $$x = 1$$ Substitute $x=1$ into $y = x - 2$: $$y = 1 - 2 = -1$$ Solution is $(1, -1)$. --- 6. Solve by graphing: $$\begin{cases} y = -2x - 1 \\ x + y = 3 \end{cases}$$ Equation 1 slope $m=-2$, y-intercept $b=-1$. Rewrite Equation 2: $$y = 3 - x$$ Slope $m=-1$, y-intercept $b=3$. Set equal: $$-2x - 1 = 3 - x$$ $$-2x + x = 3 + 1$$ $$-x = 4$$ $$x = -4$$ Substitute $x=-4$ into $y = -2x - 1$: $$y = -2(-4) - 1 = 8 - 1 = 7$$ Solution is $(-4, 7)$.