Question

consider the following system of inequalities. which of the ordered pairs are solutions to the system, and which ordered pairs are not solutions to the system? select each ordered pair in the appropriate column of the table.
\\(\

$$\begin{cases}6x + 4yleq12\\y>\frac{1}{3}x - 2\\end{cases}$$

\\)

solutions	not solutions
\\(\square(3,-6)\\)	\\(\square(3,-6)\\)
\\(\square(0,0)\\)	\\(\square(0,0)\\)
\\(\square(-3,-3)\\)	\\(\square(-3,-3)\\)
\\(\square(-4,2)\\)	\\(\square(-4,2)\\)
\\(\square(6,-2)\\)	\\(\square(6,-2)\\)
\\(\square(4,6)\\)	\\(\square(4,6)\\)

Explanation:

Step1: Check for $( - 2,6)$

For $6x + 4y\leq12$, substitute $x=-2$ and $y = 6$: $6(-2)+4\times6=-12 + 24=12$, and $12\leq12$ holds. For $y>\frac{1}{3}x - 2$, substitute $x=-2$ and $y = 6$: $6>\frac{1}{3}(-2)-2=-\frac{2}{3}-2=-\frac{2 + 6}{3}=-\frac{8}{3}$, which holds. So $(-2,6)$ is a solution.

Step2: Check for $(3,-6)$

For $6x + 4y\leq12$, substitute $x = 3$ and $y=-6$: $6\times3+4\times(-6)=18-24=-6$, and $-6\leq12$ holds. For $y>\frac{1}{3}x - 2$, substitute $x = 3$ and $y=-6$: $-6>\frac{1}{3}\times3-2=1 - 2=-1$, which is false. So $(3,-6)$ is not a solution.

Step3: Check for $(0,0)$

For $6x + 4y\leq12$, substitute $x = 0$ and $y = 0$: $6\times0+4\times0=0$, and $0\leq12$ holds. For $y>\frac{1}{3}x - 2$, substitute $x = 0$ and $y = 0$: $0>\frac{1}{3}\times0-2=-2$, which holds. So $(0,0)$ is a solution.

Step4: Check for $(-3,-3)$

For $6x + 4y\leq12$, substitute $x=-3$ and $y=-3$: $6(-3)+4(-3)=-18-12=-30$, and $-30\leq12$ holds. For $y>\frac{1}{3}x - 2$, substitute $x=-3$ and $y=-3$: $-3>\frac{1}{3}(-3)-2=-1 - 2=-3$, which is false. So $(-3,-3)$ is not a solution.

Step5: Check for $(-4,2)$

For $6x + 4y\leq12$, substitute $x=-4$ and $y = 2$: $6(-4)+4\times2=-24 + 8=-16$, and $-16\leq12$ holds. For $y>\frac{1}{3}x - 2$, substitute $x=-4$ and $y = 2$: $2>\frac{1}{3}(-4)-2=-\frac{4}{3}-2=-\frac{4 + 6}{3}=-\frac{10}{3}$, which holds. So $(-4,2)$ is a solution.

Step6: Check for $(6,-2)$

For $6x + 4y\leq12$, substitute $x = 6$ and $y=-2$: $6\times6+4\times(-2)=36-8 = 28$, and $28>12$, which is false. So $(6,-2)$ is not a solution.

Step7: Check for $(4,6)$

For $6x + 4y\leq12$, substitute $x = 4$ and $y = 6$: $6\times4+4\times6=24 + 24=48$, and $48>12$, which is false. So $(4,6)$ is not a solution.

Answer:

Solutions	Not Solutions