How to solve simultaneous inequalities? I am doing multivariable calculus, and specifically double integrals. I am facing difficulties finding the domain of the integal, however i am given the following equations:
$$1 ≤ 2x+y ≤ 2$$
$$0 ≤ x-2y ≤ 1$$
Through these two equations i am supposed to find the area of integrals for each of the variables i.e $x$ and $y$
I set them as simultaneous inequalities but it doesn't seem to help because i get the boundary for $x$ to be $2/5 ≤ x ≤ 1 $ and for $y$ to be $4/5 ≤ y ≤ 0 $ which is obviously WRONG because how can the lowest boundary for possibly have higher value than that in the higher. 
Guys, I really appreciate you help, it means the world to me.
 A: Unless your integrand looks particularly nice in your current $x,y$ coordinates, I think the sane way to do this would be to switch coordinates to
$$ u = 2x+y \\ v = x-2y $$
such that your area of integration becomes a nice axis-parallel rectangle in $u,v$ coordinates.
(You then just need to remember to divide by (the absolute value of) $\left|\begin{matrix} 2 & 1\\ 1 & -2 \end{matrix}\right|$ to take into account that the area of integration measures more in $uv$ coordinates than in $xy$ coordinates, of course).
A: Because you said you are doing double integral, if you want to keep using $x$ $y$ in the integral, then you may need to following clumsy way.
We have
$$1-2x \le y \le 2-2x$$
and
$$\frac{x-1}{2} \le y \le \frac{x}{2}$$
Now for $x>1$, we have
$$2-2x < \frac{x-1}{2}$$
and hence there is no solution.
For $x<2/5$, we have
$$\frac{x}{2}<1-2x$$
and so there is also no solution.
For $4/5\le x\le 1$, we have $2-2x \le x/2$ and $1-2x \le (x-1)/2$, hence
$$\frac{x-1}{2} \le y \le 2-2x$$
For $3/5 \le x \le 4/5$, we have $x/2 \le 2-2x$ and $1-2x \le (x-1)/2$, hence
$$ \frac{x-1}{2} \le y \le \frac{x}{2}$$
For $2/5 \le x \le 3/5$, we have $x/2 \le 2-2x$ and $(x-1)/2 \le 1-2x$, hence
$$1-2x \le y \le \frac{x}{2}$$
So your integral
$$\iint_D f(x,y)dydx=\int_{4/5}^1 \int_{(x-1)/2}^{2-2x}f(x,y)dy dx$$
$$+\int_{3/5}^{4/5}\int_{(x-1)/2}^{x/2}f(x,y)dydx+\int_{2/5}^{3/5}\int_{1-2x}^{x/2}f(x,y)dydx$$
