SOLUTION: I need some help with this word problem. Any help would be greatly appreciated. Thanks.
A photo is 3 inches longer that it is wide. A 2-inch border is placed around the ph
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Question 144909: I need some help with this word problem. Any help would be greatly appreciated. Thanks.
A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo?
Found 2 solutions by ankor@dixie-net.com, jojo14344:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo?
:
Let x = the width of the photo
:
It says,"A photo is 3 inches longer that it is wide. ", therefore:
(x+3) = the length of the photo
:
The dimensions of the whole thing: (x+3)+ 4, x + 4, which is
(x+7) by (x+4)
:
Given the area of the whole thing:
(x+7)(x+4) = 108
FOIL
x^2 + 11x + 28 - 108 = 0
:
x^2 + 11x - 80 = 0
Factor
(x+16)(x-5) = 0
The positive solution:
x = 5" is the width of the photo
and
5 + 3 = 8" is the length of the photo
:
:
Check solution)
(8+4) * (5+4) = 108
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
Sure! Let's start with the unknown ---- width = "x".
And we know the length is 3 inches longer, so length= "x+3" .
Now you attached 2-inch border around it, meaning you extended the orig. width 2inches on the left side and 2 inches on the right side right? So the new width dimension will be "x+2+2"= "x+4". Likewise the length is extended 2 inches on the top and 2 inches on the bottom, so it's new dimension will be "x+3+2+2" = "x+7".
Here we gonna use the area given with the photo and the border plus the new dimensions right? So going back in finding the area of rectangle,
A= (x+4)(x+7) ------------ eqn 1
108= x^2 + 11x + 28
x^2 + 11x -80 = 0
so (x+16)(x-5) right?
It has 2 values, x= -16 which we can't use being negative
x= 5... perfect! This the one to use.
Therefore going back to orig. dimension we can get the size of the photo, where the width = x= 5 inches. Also the length = x+3=5+3= 8inches.
In doubt? Go back eqn 1
A= (5+4)(5+7)
108in^2 = 9*12
108in^2 = 108in^2
thank you,
Jojo
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