Question

I get stuck whit these I don’t get how to do them

Answer 1

The solution to the equations is where the graphs of the two functions intersect. We can't trust the graph because it's not clear. We have to solve the system of inequalities. Let's solve for x in the first equation:
$y \leq 2x-2$
$y +2 \leq 2x$
$\frac{y+2}{2} = x$

Plug that into the second equation for x:
$y \leq ( \frac{y+2}{2})^2-3(\frac{y+2}{2})$
$\frac{y^2+4y+4}{4}- \frac{3y-2}{2} = \frac{y^2+4y+4}{4}- \frac{6y-4}{4}$
$y \leq\frac{y^2-2y}{4}$ 
$0 \leq y^2-2y-4y$
$0 \leq y(y-6)$
$6 \leq y$

Now, plug this back into the first equation to solve for x:
$6 \leq 2x-2$
$8 \leq 2x$
$4 \leq x$

So, the solution to that system is (4, 6) which looks like the D.