how do i graph .21x + .23y Less than or equal to 75

Question

raden · Answer

first, let that's an equation of straight line :
|dw:1365991228122:dw|
with (0,a) is y-intercept and (0,b) is x-intercept

raden · Answer

can u determine the values of a and b resfectively ?

anonymous · Answer

isnt it .21 and .23?

raden · Answer

no,
to get the y-intercept take x = 0 then solve for y u'll get the value of a
while to get the x-intercept take y = 0 then solve for x u'll get the value of b

raden · Answer

.21x + .23y = 75
if x = 0 then .021(0) +.23y = 75
.23y = 75
y = 75/.23 = 7500/23
so, as y-intercept at point (0, 7500/23)

e.mccormick · Answer

What RadEn is saying is perfectly correct.  Here is just another way to think about it. 

First, solve for y.  That gets it to the line equation you may be more familliar with and you can do what RadEn is talking about.  You graph the line because this is $\le$. If it was just < or >, the line would be dashed because it is not part of the answer.  

Then shade the proper side of it.  However, be careful about the rules for solving inequalities.  If you multiply or divide by a negative, the direction of the inequality changes!

anonymous · Answer

it should be shaded above?

e.mccormick · Answer

Depends on what you get for an equation, if x>y or xy, below. And x

e.mccormick · Answer

If you leave it as x+y, it can be hard to see the relationship.  That is why I like to solve for y first.

anonymous · Answer

okay but it started at less than or equal to now im confused at what way i shade

e.mccormick · Answer

$.21x + .23y\le75$
$.21x + .23y-.32y\le75-.32y$
$.21x \le75-.32y$
$.21x-75 \le75-75-.32y$
$.21x-75 \le-.32y$
$(.21x-75)/(-.32) \ge-.32y/(-.32)$
$(.21x-75)/(-.32) \ge y$

e.mccormick · Answer

Do you see where I divided by a negative and the direction changed?

anonymous · Answer

yes

e.mccormick · Answer

That final formula is what I would graph.

raden · Answer

alternative :
take a point (almost it is (0,0)), then subtitute to system inequality above :
.21x + .23y <= 75
.21(0,0) + .23(0,0) <= 75
0 <= 75

this is true, so the shaded area would forward to point (0,0). in other words it is under that line

anonymous · Answer

Thank you!

e.mccormick · Answer

RadEn has another great point there!  Testing parts of the graph aove and below the line for proper results is a great idea.  Can clear up where to shade in seconds.

e.mccormick · Answer

Once you finish your graph, here is one for comparison.

Whichever way you like, RadEn's or mine, is fine.  They both work and that is the thing about math.  He is going right for intercepts and the line, then finishing with a test point.  I am putting the line in a format I am more comfortable with and using my understanding to decide where to shade after that.   Everybody needs to understand math the way that works for them and get valid answer so they can use it properly.

anonymous · Answer

thank you!