SOLUTION: determine the point of intersection of: a) 1/2(x-2)-1/4(3-y) ≥ 2 3x-4y ≥ 2 b) x-3(y-3) ≤ 3 2x - 1/3(3-3y) ≥ 4 x +

Algebra ->  Graphs -> SOLUTION: determine the point of intersection of: a) 1/2(x-2)-1/4(3-y) ≥ 2 3x-4y ≥ 2 b) x-3(y-3) ≤ 3 2x - 1/3(3-3y) ≥ 4 x +       Log On


   



Question 692801: determine the point of intersection of:
a) 1/2(x-2)-1/4(3-y) ≥ 2
3x-4y ≥ 2
b) x-3(y-3) ≤ 3
2x - 1/3(3-3y) ≥ 4
x + 5/3y ≥ 1

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
a)
1%2F2%28x-2%29-1%2F4%283-y%29+%3E=+2
3x-4y+%3E=+2
_________________
x%2F2-2%2F2-3%2F4-y%2F4+%3E=+2

4x%2F2-4%2A2%2F2-4%2A3%2F4-4%2Ay%2F4+%3E=+2%2A4
2x-4-3-y+%3E=+8
2x-4-3-8+%3E=+y
2x-15+%3E=+y
y%3C=2x-15


3x-4y+%3E=+2
3x-2+%3E=+4y
%283%2F4%29x-1%2F2+%3E=+y
y%3C=%283%2F4%29x-1%2F2
first graph both lines as y=2x-15 and y=%283%2F4%29x-1%2F2, find x and y-intercept and find intersection point
y=2x-15...set x=0 and find y
y=2%2A0-15........=>.........y=-15; y-intercept is at (0,-15)

y=2x-15...set y=0 and find x
0=2x-15........=>....x=15%2F2=>....x=7.5; x-intercept is at (7.5,0)
--
y=%283%2F4%29x-1%2F2...set x=0 and find y
y=%283%2F4%290-1%2F2........=>.........y=-1%2F2; y-intercept is at (0,-1%2F2)

y=%283%2F4%29x-1%2F2...set y=0 and find x
0=%283%2F4%29x-1%2F2........=>....%283%2F4%29x=1%2F2=>....x=%281%2F2%29%2F%283%2F4%29=4%2F6=2%2F3; x-intercept is at (2%2F3,0)
draw a lines, note both of them are a part of solution
+graph%28+600%2C+600%2C+-10%2C+20%2C+-20%2C+20%2C%283%2F4%29x-1%2F2%2C+2x-15%29+
now shade the part where y%3C=2x-15 and y%3C=%283%2F4%29x-1%2F2


b)
do same with these inequalities

+x-3%28y-3%29+%3E=+3....=>...+x-3y-9+%3E=+3=>...+x-3-9+%3E=+3y=>...+x-12+%3E=+3y...=>...%281%2F3%29+x-4+%3E=y...=>...y%3C=%281%2F3%29+x-4

2x+-+1%2F3%283-3y%29+%3E=+4..=>..2x-1-y+%3E=+4..=>..2x-1-4+%3E=+y..=>..2x-5%3E=+y..=>..y%3C=2x-5

+x+%2B+5%2F3y+%3E=+1...=>...+3x+%2B+5y+%3E=+3...=>...5y%3E=-3x%2B3..=>...y%3E=-%283%2F5%29x%2B3%2F5
same way you find x and y-intercepts and draw a lines than shade the area that belongs to a solution ; it will look like this: