How do you graph the inequality –3x – 4y<=12?

1 Answer
Feb 5, 2015

To graph this kind of inequality, the best thing is to manipulate them and obtain a relation of the form $y \setminus \le f \left(x\right)$ (or $y \setminus \ge f \left(x\right)$).

In fact, we know that $y = f \left(x\right)$ is exactly the graph of $f$, and so $y \setminus \le f \left(x\right)$ (or $y \setminus \ge f \left(x\right)$) represents all the portion of the plan below (or above) the graph of $f$.

Let's do those manipulations: starting from
$- 3 x - 4 y \setminus \le 12$,
adding $4 y$ at both sides we get
$- 3 x \setminus \le 4 y + 12$.
Subtracting 12 at both sides, we have
$- 3 x - 12 \setminus \le 4 y$.
Dividing both sides by $4$ we finally have
$- \frac{3}{4} x - 3 \setminus \le y$
which can of course be read as
$y \setminus \ge - \frac{3}{4} x - 3$

We thus have $f \left(x\right) = - \frac{3}{4} x - 3$, which is a line and so it's very easy to graph. Once graphed, you need to consider all the portion of plan above the line to solve your inequality.

Here's the graph: graph{-3x-4y \le 12 [-10, 10, -5, 5]}