Question

Hi! My teacher wanted me to sketch the graph of solution set of each system. For the first question, she wants me to use the slope intercept form and the second question using a table of values, but, I'm a little lost. Can someone help? Thanks!


1.) 3x + 2y < 6
x > 0
y > 0


2.) 2x2 + y ≥ 2
x ≤ 2
y ≤ 1

Answer 1

3x + 2y < 6
2y < -3x + 6
y < -(3/2)x + 3
so the slope is -3/2 and the y-intercept is 3
so start with the y-intercept of 3 , or the point (0,3)
slope is rise/run = -3/2
so from (0,3) , drop down 3 , then 2 units to the right. This should bring you to the point (-3,5)
Joint the points with a dotted line, shade in everthing in the first quadrant below the dotted line
see: http://www.wolframalpha.com/input/?i=plot+3x+%2B+2y+%3C+6+%2C+y+%3E+0+%2C+x+%3E+0
Lines should be dotted


2x^2 + y ≥2
y ≥ -2x^2 + 2

shade in everything "above" the graph of
y = -2x^2 + 2 , making the boundary a solid line

Restrict your graph to all values of x ≤ 2 , and y values ≤ 1

here are some points of the parabola boundary

(0,2)
(1,0), (-1,0)
(2,-2) , (-2,-2)

etc.