Describe and graph the solution region of 3|x| + 4|y| <= 12.

Step-by-step solution

1. 1

   Exploit the double symmetry first. Replacing $x$ by $-x$ or $y$ by $-y$ leaves $3|x|+4|y|$ unchanged, so the region is symmetric about both axes. It is therefore enough to understand the first quadrant and reflect twice.

2. 2

   Find the $x$-intercepts. Set $y=0$:

   $3|x|\le 12\;\Longrightarrow\;|x|\le 4\;\Longrightarrow\;-4\le x\le 4$

   so the boundary meets the $x$-axis at $(\pm 4,0)$.

3. 3

   Find the $y$-intercepts. Set $x=0$:

   $4|y|\le 12\;\Longrightarrow\;|y|\le 3\;\Longrightarrow\;-3\le y\le 3$

   giving $(0,\pm 3)$.

4. 4

   Recognise the boundary as four line segments. In the first quadrant the equality $3x+4y=12$ is a straight line, and the symmetry copies it into the other three quadrants. Four straight edges joining $(4,0)$, $(0,3)$, $(-4,0)$, $(0,-3)$ form a diamond (a rhombus) centred at the origin — the absolute values are what replace a smooth ellipse with straight edges.

5. 5

   Decide which side to shade. Test the origin: $3|0|+4|0|=0\le 12$ $\checkmark$, so the origin is included and the region is the inside of the diamond. The inequality is non-strict, so the four edges are drawn solid and belong to the solution.

6. 6

   Check a boundary point and an outside point. $(2,\tfrac32)$: $6+6=12\le 12$ $\checkmark$ on the edge. $(4,1)$: $12+4=16\not\le 12$ $\checkmark$ correctly outside.