How do you graph the region $3x-4y>= -12$ ?

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Answer 1 · 2015-02-10T01:42:15

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Since $3x - 4y = -12$ is linear, the easiest place to start is to draw the line for that equation by determining the x and y axii intercepts.

When $y = 0$ then $3x - 4y = -12$ becomes $3x = -12$ or $x = -4$

When $x = 0$ then $3x - 4y = -12$ becomes $-4y = -12$ or $y = 3$

We can now draw the line for the equation $3x - 4y = -12$ by drawing a straight line through the two points $(-4,0)$ and $(0,3)$

All that remains is to determine which side of that line is represented by $3x - 4y = -12$

Consider the point $(x,y) = (0,0)$ and apply it to the given expression
$3x - 4y >= -12$ giving $3(0) - 4(0) >= -12$ or $0 >= -12$

Since this is obviously true $(x,y) = (0,0)$ must be within the area described by $3x - 4y >= -12$

The required graph is therefore the unshaded area in the diagram (plus the line for $3x - 4y = -12$ ).

How do you graph the region 3x-4y>= -123x-4y>= -12?