# Solve each system by graphing...?

• Jun 15th 2008, 09:05 PM
AlphaRock
Solve each system by graphing...?
What does it mean when a question asks: "Solve each system by graphing?"

Then:

A) y > .5x - 2
3x + 4y<6

B) 4x-y <= (less than or equal to) 8
y <= (less than or equal to) 5
x + 3y >6

I'm confused as to how to graph these and why it should be solved like that.

I understand the steps are:

1. Isolate for y. (I can't remember about the rule where it switches from > to <. If anybody remembers, please let me know!)
2. Graph.
3. Color it in. < Confusing...
• Jun 15th 2008, 09:14 PM
kalagota
when you do your first step, do it with an "=" first.. just ignore > or <..

for the second one, if you see $<$ or $>$, instead of a solid line, draw a dashed line.. "<- - - - - - - - ->"; however, if you see $\leq$ or $\geq$, draw the solid one..

for your third one, just pick points not on your lines, substitute it to your system of equations and shade that part where your point belongs if it satisfies the inequalities.. if not, pick another points from other part of the plane..
• Jun 15th 2008, 09:29 PM
earboth
Quote:

Originally Posted by AlphaRock
What does it mean when a question asks: "Solve each system by graphing?"

Then:

A) y > .5x - 2
3x + 4y<6

...

1. Isolate for y. (I can't remember about the rule where it switches from > to <. If anybody remembers, please let me know!)
2. Graph.
3. Color it in. < Confusing...

Each inequality defines a half plane in the coordinate plane which is bounded by a straight line.

1. Solve the inequalities for y:

$y > \frac12 x - 2$ the lower bounds of this half plane is the line with the equation $y = \frac12 x - 2$

$3x+4y<6~\iff~y<-\frac34 x + \frac32$ the upper bounds of this half plane is the line with the equation $y = -\frac34 x + \frac32$

In both cases the bounds don't belong to the half planes in question.

2. Draw the lines.
Shade the area above the first line and shade the area below the second line.
The solution consists of all ordered pairs (x, y) which can be drawn as points in the brown area because they are simultaneously above the first line and below the second line
• Jun 15th 2008, 10:07 PM
earboth
Quote:

Originally Posted by AlphaRock
What does it mean when a question asks: "Solve each system by graphing?"
...

B) 4x-y <= (less than or equal to) 8
y <= (less than or equal to) 5
x + 3y >6

...

1. Transform the inequalities:
$4x-y\leq 8~\implies~ y\geq 4x-8~\text{ draw the line } y= 4x-8$ To the solution belong all points above the line including the line.

$y\leq 5~\text{ draw the line } y= 5$ To the solution belong all points below the line including the line.

$x+3y>6~\implies~ y >-\frac13 x+2~\text{ draw the line } y= -\frac13 x+2$ To the solution belong all points above the line but not the points on the line.