# How do I solve for 2x+3y>-9 and 4x-3y<-9 ?

See below:

#### Explanation:

Let's first work with the line boundaries and find the point of intersection, then deal with shaded area.

To work with the lines, let's change the inequalities into equalities:

$2 x + 3 y = - 9$
$4 x - 3 y = - 9$

We can add the two equations together (which will eliminate the $y$ terms):

$6 x = - 18$

$x = - 3$

which means:

$2 \left(- 3\right) + 3 y = - 9$

$- 6 + 3 y = - 9$

$3 y = - 3$

$y = - 1$

Here are the two lines graphically and they intersect at $\left(- 3 , - 1\right)$:

graph{(2x+3y+9)(4x-3y+9)=0}

Looking at the graph, let's call the regions East (which has the origin), North, West, and South. Which region(s) get shaded?

Let's first convert the general form of the lines to slope-intercept:

$2 x + 3 y = - 9 \implies y = - \frac{2}{3} x - 3$
$4 x - 3 y = - 9 \implies y = \frac{4}{3} x + 3$

Now we can better identify which line is which (the $y$-intercepts show which line is which - the first equation is the line headed "down" from left to right and the second equation is the line headed "up" also from left to right).

For the first inequality:

$2 x + 3 y > - 9$

the origin is a valid solution:

$2 \left(0\right) + 3 \left(0\right) > - 9 \implies 0 > - 9$

so that's regions East and North.

For the second inequality:

$4 x - 3 y < - 9$

the origin is not a valid solution:

$4 \left(0\right) - 3 \left(0\right) < - 9 \implies 0 < - 9$

so that's regions North and West that are valid.

This is an "and" question, meaning we want the regions to be valid for both expressions, so our region is North.

The final graphical answer is therefore the line graph above, with dotted lines (to represent that the points on the lines themselves are not part of the solution set) with the shaded area (the area with the valid points) being the North area (which contains, among other points, $\left(- 3 , 0\right)$