last year, the area of a garden is 60 sq ft.?

Question

this year it is 3 ft shorter and 1 ft wider with same area. what is the length and width of last year's garden. please help. i am helping somebody too but i am stuck.

thanks

Answer 1

well ok so the total area is 60 feet. and that is equal to the length x width.
lw = 60.
now this year, the length decreased by 3 and width increased by 1 and its still same area. so the new length is (l-3) and new width is (w+1) so
(l-3)(w+1) = 60

so now rewrite the first one in terms of l and you get l=60/w so

(60/w - 3)(w+1) = 60
60 - 3w + 60/w - 3 = 60
-3w + 60/w - 3 = 0. and now multiply the equation by w

-3w^2 -3w + 60 = 0 or dividing by -3w
w^2 + w - 20 = 0
(w+5)(w-4)=0
so w = -5 or 4. obviously it cant be -5, so w=4. which means the length = 60/4 = 15.

so last years garden was 15 x 4

Answer 2

L ,W are last year's length and width.

LW = 60 so L = 60/W

but also (L - 3)(W + 1) = 60

or LW - 3W + L - 3 = 60 But LW = 60 so L - 3W - 3 = 0

Then L = 3W + 3 = 60/W or 3W^2 + 3W - 60 = 0 , that is,

0 = W^2 + W - 20 = (W + 5)(W - 4)= 0 thus W = 4 or W = -5

But w = -5 does not make sense in this problem, so W = 4

and L = 60/W = 15.

Last year the garden was 4 ft by 15 ft

This year it is 5 ft by 12 ft

Answer 3

Brute force solution ...

list the possible factors ..
1,60
2,30
3,20
4,15
5,12
6,10

test ... take away 3 from one factor, add 1 to the other and see what happens ...

bad news for the first three on the list.

then, 4, 15 ... 15 = 12, and
4+1 = 5

so it works. 4 and 15.

Using algebra ...

xy=60
(x+1)* (y+3) = 60

sub in ... x=y/60

solve for y ...

you get 4,15

Answer 4

Last year's garden was 15 feet by 4 feet.

Answer 5

well there are 2 ways to solve do a predict & test chart or solve it algerbraicly