Solve the following system of linear inequalities by graphing.

x-3y>6
3x+2y>12

For the first equation I came up with y<1/3x+-2 (I flipped the sign because I divided by a negative)

The second equation, y>-3/2x+6

For the graph: could someone explain in detail what side of the line I graph? Thanks!!!!

Plot those lines (y=1/3x -2, and the othera, making the lines dashes. Your solution is the area which is above the second line, and below the first.

To graph the system of linear inequalities, we need to draw the boundary lines for each inequality and shade the appropriate regions based on the given inequality conditions.

For the first inequality x - 3y > 6:

1. Let's rewrite the equation to find the slope-intercept form (y = mx + b).
x - 3y > 6
-3y > -x + 6
Divide both sides by -3, and remember to reverse the inequality sign:
y < (1/3)x - 2

2. Start by graphing the boundary line, which is the equation y = (1/3)x - 2.
To do this:
- Choose two x-values, such as x = 0 and x = 3.
- Plug them into the equation to find the corresponding y-values:
For x = 0, y = (1/3)(0) - 2 = -2
For x = 3, y = (1/3)(3) - 2 = -1
- Plot the points (0, -2) and (3, -1), and draw a line through them.

3. Now, determine which side of the line to shade. Since the inequality is y < (1/3)x - 2, we shade below the line.

For the second inequality 3x + 2y > 12:

1. Solve the equation for y to obtain the slope-intercept form:
3x + 2y > 12
2y > -3x + 12
Divide both sides by 2 and reverse the inequality sign:
y > (-3/2)x + 6

2. Graph the boundary line, which is the equation y = (-3/2)x + 6.
Following similar steps as before, choose two x-values, substitute them into the equation, plot the corresponding points, and draw a line through them.

3. The inequality is y > (-3/2)x + 6, so we shade above the line.

Finally, shade the appropriate regions for each inequality on the graph, considering the directions discussed earlier. The region that is shaded by both inequalities will be the solution to the system of linear inequalities.