Algebra two inequality

Question

Graph the area bounded by

 y< 1/2x+6, X+3y>=12, X>=0 and x<=12

 

part i. graph the bounded by X>=0 and x<=12. shade the region that satisfies both inequalites 

 

part ii. to graph you've already produced, add the region bounded by the lines y<1/2x+6 and x+3y>=12. shade the region that satisfies both of these inequalties and the ones in part I. 

 

Answer 1

You can graph at DESMOS dot Com

For now, I will find the intersection point :

y = (1/2)x + 6 vs x + 3y = 12

Substitution:

x + 3( (1/2) x + 6) = 12

x + (3/2)x + 18 = 12

(2/2)x + (3/2)x + 18 = 12

(5/2)x + 18 = 12

(5/2)x = -6

x = (-6)(2/5) = -12/5

then y = (1/2)(-12/5) + 6

= (-12/10) + 6

= (-6/5) + 6

= (-6/5) + 30/5

= 24/5

check: x=-12/5 and y = 24/5

plugs into second equation:

x + 3y = -12/5 + 3(24/5) =

-12/5 + 72/5 = 60/5 = 12

yes it checks !!

the lines meet at (x=-12/5 , y =24/5)

tests the origin (x=0,y=0) in the original inequalities

1st inequality

0<(1/2)(0) + 6 --> 0<6 is true, so the origin is INCLUDED in the shading of the graph.

intercept is (0,6); two other points are (2,7) and (4,8) ; per the slope y increases by 1 while x increases by 2

connect those points with a DOTTED/DASHED line and shades towards the origin in BLUE

2nd inequality

0 + 3(0) > 12 ---> 0>12 is FALSE, so thee origin is NOT INCLUDED in the shading of the graph

two other points are (3,3) and (0,4) is the intercept

connects those points with a SOLID line and shades AWAY from the origin in YELLOW

the solution will appear in GREEN