How do you graph y <2x+4 and -3x-2y>=6?

1 Answer
Jun 1, 2018

See explanation

Explanation:

Given:

y<2x+4" ".......................Eqn(1)

-3x-2y>=6" ".....................Eqn(2)

color(blue)("Converting "Eqn(2)" into standardised form ")

Consider Eqn(2)

Add 2y to both sides and subtract 6 from both sides giving:

-3x-6>=2y

Divide both sides by 2

-3/2 x-6/2>=y

Write the order as per convention

y<=-3/2x-3" "..........Eqn(2_a)
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Plot them color(magenta)(ul("as if they were"))
y=-3/2 x-3 color(white)("dddd") ->color(white)("dddd")"really is: " y<2x+4 " "Eqn(2_a)

Feasible solution area is below color(white)()"and "ul(color(red)("including the solid line"))color(white)("d") Eqn(2_a)

and:

y=2x+4 color(white)("dddd") ->color(white)("dddd")"really is: " y<2x+4" "Eqn(1)

Feasible solution area is below and
color(red)(ul("excluding the dotted line "))Eqn(1)

The solution area is where these two are coincidental (coincide).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine key point - Intersection of the two lines")

y=2x+4 " "............Eqn(1)
y=-3/2 x-3" "......Eqn(2_a)

Eqn(1)-Eqn(2) to 'get rid' of the y's

0=7/2x+7

color(green)(x=-7xx2/7=-2)

Substitute x=-2 in Eqn(2_a) giving:

y=(-3/2)(-2)-3 = +3-3=0

color(blue)(ul(bar(| color(white)(2/2) "Intersection"->(x,y)=(-2,0)color(white)(2/2) |))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine key point - axis intercepts for "y<2x+4)

Set y=2x+4

Set x=0 -> color(blue)(bar(ul(|color(white)(2/2)y_("intercept")=4color(white)(2/2)|))

Set y=0 -> color(blue)(ul(bar(|color(white)(2/2) x_("intercept")=-2color(white)(2/2)|))
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Determine key point - axis intercepts for "-3x-2y>=6)

Eqn(2_a)->y<=-3/2x-3

Set y=-3/2x-3

Set x=0 ->color(blue)( ul(bar(| color(white)(2/2)y_("intercept")= -3color(white)(2/2)|)))

Set y=0->color(blue)( ul(bar(| color(white)(2/2)x_("intercept")= -2color(white)(2/2)|)))

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