Question

How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?

Answer 1

\[\int f(x)-g(x)dx\]i believe

Answer 2

\[\left| 3x-18 \right|+\left| 2y+7 \right|\le3\]

Answer 3

or triangles :)

Answer 4

That looks pretty difficult to graph...

Answer 5

looks can be deceiving ....

Answer 6

|3x-18| + |2y+7| - 3 <= 0

Answer 7

better try this on the wolf

Answer 8

http://www.wolframalpha.com/input/?i=%7C3x-18%7C+%2B+%7C2y%2B7%7C+-+3+%3C%3D+0

Answer 9

what methods for area can we use? vectors, determinants, simpler stuff?

Answer 10

Well, this is for the AMC12, where the highest level math needed is Trig/Precal and there are no calculators allowed, so is there anyway I can draw this by hand? Or find the area algebraically?

Answer 11

algebra most likely; we know when x=6 the the left || drops out

Answer 12

|3x-18| + |2y+7| <= 3 ; x=6

|2y+7| <= 3

2y+7 <=3 or -2y-7 <= 3
2y <= -4 2y +7 >= -3
y <= -2 2y >= -10
y>= -5

^
|
|
| -2
|
|
| -5
|
v
y

that might be useful .....

Answer 13

then we do the same when we drop the y side

|3x-18| <= 3

3x-18 <= 3 or 3x-18 >= -3
3x <= 21 3x >= 15
x <=7 x >= 5



^
<------|---------------> x
| 5 7
| -2
|
|
| -5
|
v
y

just an idea, still dont know if its gonna be helpful tho

Answer 14

ohh, thtas actually what the graph of the wplf gives;

Answer 15

if i could type that would read more impressive lol

Answer 16

x=6 is a vertex; and y= -7/2 is a vertex

(6,-2), (6,-5), (5,-7/2),(7,-7/2) are the points of that tilted square thing

Answer 17

wait, so how did out find the vertices (btw, great work, it's really easy to follow)

Answer 18

once you know the vertices, the points, you can do tests to see if its right angled and square or if its more trapazoidal and such

Answer 19

i zeroed out each |...| by setting its variable to something that would make it zero

Answer 20

then i solved for the remaining |....| and its results are the other parts of our orered pairs

Answer 21

i think i have something easier

Answer 22

im sure there is an easier way, but thats just not how a roll :)

Answer 23

ohhh, ok, thanks amistre64 for your help!

Answer 24

youre welcome :)

Answer 25

well, I have the solution pamplter and there is a solution (and it was multiple choice
Choices:
A) 3
b) 7/2
c)4
d) 9/2
e) 5
)

Answer 26

well we can go through all those given and check to see if the inequality holds
this way would be much much easier :)

Answer 27

check x=6, y=-3 as a solution

Answer 28

there is an area of possible created by zeroing the x part and gettting y values; then zeroing the y part and getting x values

Answer 29

those 4 points of interest create parallelgram or some sort of shape that confinces the solution set

Answer 30

(6,-2), (6,-5), (5,-7/2), (7,-7/2)
-6+2 -6+2 -6+2 -6+2
---------------------------
(0,0) (0,-3) (-1,-3/2) (1,-3/2)

are what we look like when we set one vertice to the origion

Answer 31

if I were to raise it another 3/2 we could have it centered at the origin

(0,0) (0,-3) (-1,-3/2) (1,-3/2)
+3/2 +3/2 +3/2 +3/2
-----------------------------
(0,1.5) (0,-1.5) (-1,0) (1,0)

Answer 32

well, it aint a square, but its alteast rhombused

Answer 33

my next idea, brilliantly exectuted lol is to find the area of just one of those triangle and the 4x it

Answer 34

height = 1.5
base = 1

A = bh/2 = 1.5/2 = .75 right?

.75*4 = total area

Answer 35

im going with 3 :)