# How do you graph the inequality x + 2y <=8, x>=0 and y>=0?

Mar 6, 2018

as in the explanantion

#### Explanation:

For the blue line: $x + 2 y \le 8$

Write as: $x + 2 y = 8 \textcolor{red}{\leftarrow \text{ For only the line}}$

Subtract $x$ from both sides

$2 y = - x + 8$

Divide both sides by 2

$y = - \frac{1}{2} x + 4$
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$\textcolor{g r e e n}{\text{As we are using "<=" the line is solid}}$

IFF we had been using $<$ the line would be dotted or dashed

So as we have not 'moved' the $y$ or multiplied by (-1) the sign $\le$ is the same way round in relation to $y$ giving:

$y \le - \frac{1}{2} x + 4$ giving us the acceptably reagion for THIS LINE ALONE.

The other conditions of $x > 0 \mathmr{and} y > 0$ means that we can only use the positive region of $y \le - \frac{1}{2} x + 4$