SOLUTION: Solve the problem: The printed matter on a 12 by 18 centimeter page of a book must cover 40 square centimeters. If all margins are to be the same width, how wide should the mar

Question 4796: Solve the problem:
The printed matter on a 12 by 18 centimeter page of a book must cover 40 square centimeters. If all margins are to be the same width, how wide should the margins be?
Found 2 solutions by Earlsdon, Abbey:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
It's often helpful to draw yourself a diagram that graphically depicts the situation to be solved.
I don't know how to draw diagrams with this system but you can readily see that your picture would show a smaller rectangle centered inside of a larger rectangle. The smaller rectangle represents the area (40 cm^2) covered by the printed matter while the larger rectangle represents the page whose dimensions are 12 cm X 18 cm.
Let the width of the uniform margin be x cm.
Now you can set up your equation for finding x:
We'll need to write the equation for the area of the smaller rectangle in terms of x, the width of the margin.
The width othe smaller rectangle can be written: W = 12-2x and the length can be written: L = 18-2x
The area. A = L*W = (18-2x)(12-2x) = 40 cm^2
So as not to take all the fun out of this for you, I'll let you work this out yourself:
(18-2x)(12-2x) = 40
Hint: You will get a quadratic equation that looks like this: 4x^2-60x+176 = 0
You need to solve this for x and when you do, you will find that there are two values of x (no surprise here since solutions to quadratic equations yield two roots).
What needs to be determined is...which value of x is the solution to the problem? Both values are positive! But only one of the values makes any physical sense in this problem.
Try this. If you run into insurmountable difficulties, post another question.

Answer by Abbey(339) (Show Source): You can put this solution on YOUR website!
Let L = Length of printed matter
Let W = Width of printed matter
let M = Margin
Area of printed matter = 40 = L*W Length of page = 18cm = 2M+L
Width of page = 12cm = 2M+W
Using the elimination method:
18=2M+L
12=2M+W
Subtract the 2nd equation from the first to get
6=L-W
add w to both sides:
6+W=L
Use the substitution method on the area equation to get:
40=(6+W)*W
Clear the parentheses:

Subtract 40 from both sides to set equal to zero and solve for x:

Factor:
(w+10)(w-4)
w=-10, w=4
We can eliminate the negative because this is a distance we are discussing, so the width of the printed matter is 4cm
the length of the printed matter is 10 cm
Going back to the system of equations, we know the page length is
18=2M+L
18=2M+10
8=2M
4=M
The margin is 4 cm.
Check by adding the margins to the width and length.
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