Chapter 8 linear appliCaTionS 305

Since the original length is 121  =  23 of the original width, lw=^32. Replace
each l by^32 w.
3
2

97 3

2

43

3

2

97 3

2

(^243)
ww ww
www
+= +




• += +
  
  
  


• 
  

  ()
  ( ))
  3
  2
  97 3
  2
  9
  2
  412 3
  2
  ww^22 += ++ww+  w^2 oneachsidecancelss.
  
  
  
  =++
  =+ +=+=
  
  97 9
  2
  412
  97 17
  2
  12 9
  2
  4 9
  2
  8
  2
  17
  2
  ww
  w  
  
  −−

  
  

  ⋅=⋅

  
  

  12 12
  85 17
  2
  2
  17
  85 2
  17
  17
  2
  10
  w
  w
  w
  The original width is 10 inches and the original length is 23 w (= ^3210 ) = ^15
  inches.

  
  

3. Let r represent the original radius. Then r + 5 represents the new radius,
   and A = or^2 represents the original area. The new area is 155o square
   inches more than the original area, so 155o + A = o(r + 5)^2 = 155o + or^2.

o(r + 5)^2 = 155o + or^2
o(r + 5)(r + 5) = 155o + or^2

o(r^2 + 10r + 25) = 155o + or^2
or^2 + 10ro + 25o = 155o + or^2 (or^2 on each side cancels.)

10 ro + 25o = 155o
–25o –25o

10 r o = 130o

(^) r = 130 o
10 o
r = 13
The original radius is 13 inches.