Inequality Area

Algebra Level 4

What is the area formed by the inequality ∣ 3 x − 18 ∣ + ∣ 2 y + 7 ∣ ≤ 3 |3x-18|+|2y+7| \leq 3 ?

Find a creative way to approach the problem and post your solution! :D

3.5 3 4 4.5

Finn Hulse by Finn Hulse , 21, USA

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4 solutions

Michael Mendrin
Apr 14, 2014

If x < 5, then ∣ 2 y + 7 ∣ < 0 \left| 2y+7 \right| <0 , which is not possible. Likewise for x > 7. We use the same reasoning for y < -5 and y > -2, both of them leading to ∣ 3 x − 18 ∣ < 0 \left| 3x-18 \right| <0 Hence we have a kite of width 2 and height 3, with an area of 3. I make the presumption that the sides are straight because the inequality equation is linear.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Perfect! :D

Finn Hulse - 7 years, 1 month ago

I think you can ignore the +7 and -18 parts, and it gives you the same answer. Not a big thing, but I wanted to share this idea.

Park Sejin - 5 years, 8 months ago
Varun Gupta
Apr 30, 2014

transform the origin to (18/3, -7/2). so now inequality becomes 3|x'|+2|y'| <= 3. now required area = 4* area in 1st quadrant = 4 [area of triangle having vertices (0,0) (1,0) and (0,1.5) ] =4 [1/2(1.5*1)=3.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Jon Haussmann
Apr 21, 2014

This is from the 2008 AMC 12A .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah bro! Favorite one, to be honest. :D

Finn Hulse - 7 years, 1 month ago
Archit Wagle
Dec 29, 2014

the figure can be shifted to ( 0 , 0 ) (0,0) as it wont affect the area

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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