## anonymous 4 years ago How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?

1. amistre64

$\int f(x)-g(x)dx$i believe

2. anonymous

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

3. amistre64

or triangles :)

4. anonymous

That looks pretty difficult to graph...

5. amistre64

looks can be deceiving ....

6. amistre64

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

7. amistre64

|dw:1328570964268:dw|better try this on the wolf

8. amistre64
9. amistre64

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

10. anonymous

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?

11. amistre64

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

12. amistre64

|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 .....

13. amistre64

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

14. amistre64

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

15. amistre64

if i could type that would read more impressive lol

16. amistre64

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

17. anonymous

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

18. amistre64

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

19. amistre64

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

20. amistre64

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

21. myininaya

i think i have something easier

22. amistre64

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

23. anonymous

ohhh, ok, thanks amistre64 for your help!

24. amistre64

youre welcome :)

25. anonymous

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 )

26. myininaya

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

27. amistre64

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

28. amistre64

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

29. amistre64

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

30. amistre64

(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

31. amistre64

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)

32. amistre64

well, it aint a square, but its alteast rhombused

33. amistre64

|dw:1328573417841:dw|

34. amistre64

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

35. amistre64

height = 1.5 base = 1 A = bh/2 = 1.5/2 = .75 right? .75*4 = total area

36. amistre64

im going with 3 :)