$S$ is the set of all $(x_{1},x_{2}) \in \mathbb{R} \times \mathbb{R}$ with $x_{1}+x_{2} \geq 3$ and $-x_{1}+2x_{2}=6$
What's its implicit and explicit set?
Implicite: $S=\left\{(x_{1},x_{2})|x_{1},x_{2} \in \mathbb{R}\right\}$
For the explicite we somehow have to calculate the solutions for $x_{1}$ and $x_{2}$ (?) The problem is there is this inequality sign...
$$-x_{1}+2x_{2}=6$$
$$x_{1}=2x_{2}-6$$
Now take the other formula and replace $x_{1}$:
$$2x_{2}-6+x_{2} \geq 3$$
$$3x_{2}-6 \geq 3$$
$$x_{2} \geq 3$$
Now we need to find out $x_{1}$:
$$x_{1}+x_{2} \geq 3$$
$$x_{1} \geq 0$$
This means the explicite set would be: $S=\left\{x_{1} \geq 0, x_{2} \geq 3\right\}$
Please can you tell me, did I do it correctly? I have my doubts because none of the sets I have calculated include both the given formulas...
I don't see a better way to write this with set-builder notation than
$$\{(x_1,x_2)\in \mathbb R\times \mathbb R\mid x_1+x_2\geq 3 \text{ and } -x_1+2x_2=6\}$$
Or
$$\{(x_1,x_2)\in \mathbb R\times \mathbb R\mid x_1+x_2\geq 3 \}\cap\{(x_1,x_2)\in \mathbb R\times \mathbb R\mid -x_1+2x_2=6\} $$
As discussed in the comments, nobody yet understands how you would write this set explicitly since it has infinitely many elements.