Question 68289
Solve the following system of linear inequalities by graphing.
:
When the "<" sign is underlined, it means, Less than or equal to;
We can write it like this:
:
3x + 4y <= 12
x +  3y <= 6
x>0
y>0
:
put the equation in the slope/intercept form:
4y <= 12 - 3x
y <= 12/4 -(3/4)x
y <= 3 - (3/4)x
y <= -(3/4)x + 3 is the slope/intercept form
:
3y <= 6 - x
y <= 6/3 - (1/3)x
y <= 2 - (1/3)x
y <= -(1/3)x + 2 
:
Plot at least two points for each equation. Since there will only be positive
points (It said both x & y > 0). plot x = +1 and x = +3
:
For example x = 1
y = -(3/4)x + 3
y = -(3/4)(1) + 3
y = -3/4 + 3
y = -.75 + 3
y = + 2.25
:
x = 4
y = -(3/4)(4) + 3
y = -3 + 3
y = 0
:
Plot coordinates: 1, 2.25 and 4, 0
:
Do the same with the second equation;
:
You graph should look like this: 
{{{ graph( 300, 200, -2, 6, -2, 6, -.75x+3, -.33x+2 ) }}}
:
Note the feasibility region is above the x axis, to right of the y axis
Below or at the graph lines whichever is lower.