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.
- 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.