SOLUTION: The width of Jennica’s small painting is 3 inches less than its length. A 1-inch border is put around the painting to protect the sides of her masterpiece. The area of the painti
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-> SOLUTION: The width of Jennica’s small painting is 3 inches less than its length. A 1-inch border is put around the painting to protect the sides of her masterpiece. The area of the painti
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Question 1172184: The width of Jennica’s small painting is 3 inches less than its length. A 1-inch border is put around the painting to protect the sides of her masterpiece. The area of the painting and the border is 108 square inches. What is the length and the width of Jennica’s original painting? Found 2 solutions by ankor@dixie-net.com, greenestamps:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The width of Jennica’s small painting is 3 inches less than its length.
A 1-inch border is put around the painting to protect the sides of her masterpiece.
The area of the painting and the border is 108 square inches.
What is the length and the width of Jennica’s original painting?
:
Let L = length of original painting
let w = it's width
:
write an equation for each statement
:
" The width of Jennica’s small painting is 3 inches less than its length."
w = L - 3
;
" A 1-inch border is put around the painting to protect the sides of her masterpiece."
1 inch borders adds 2 inches to length and width
Overall dimensions: (L+2) by (w+2)
:
" The area of the painting and the border is 108 square inches."
(L+2)*(w+2) = 108
replace w with (L-3)
(L+2)*((L-3)+2) = 108
(L+2)*(L-1) = 108
FOIL
L^2 + 2L - L - 2 = 108
L^2 + L - 2 - 108 = 0
L^2 + L - 110 = 0
Factors to
(L+11)(L-10) = 0
the positive solution
L = 10 inches is the original length of the painting
the
10-3 = 7 inches is original width
:
:
Check solution with overall dimensions: 12 * 9 = 108
(1) If an algebraic solution is required, I suggest using a single variable, instead of using two variables and then doing substitution to obtain a single equation in one variable.
Furthermore, since the area given is of the painting plus border, use the length and width of the painting plus border in your equation, then subtract the border to get the answer (the dimensions of the painting without the border).
x = length (inches) (painting plus border)
x-3 = width (3 inches less than the length)
The area of the painting plus border (length times width) is 108 square inches:
OR
Select the positive solution, since negative numbers don't make sense for lengths.
The length of painting plus border is x=12 inches; the width is 9 inches.
Subtract 2 inches from each dimension to get the dimensions of the painting without the 1-inch border.
ANSWERS: The length of the painting is 12-2=10 inches; the width is 9-2=7 inches.
Note that the formal algebra didn't get you any closer to the answer -- to solve the quadratic equation, you had to find two numbers whose difference is 3 and whose product is 108. But that's what the original problem asked you to do....
So the quick and easy mental solution is to find that 12-9=3 and 12*9=108, which means the dimensions of the painting plus border are 12 by 9, making the dimensions of the painting 10 by 7.
