MHB How Do You Calculate the Dimensions of Two Equal-Area Rectangular Tables?

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Two rectangular tables are equal in area, with the first table's length being one and a half times its width (x), and the second table's length being three times its width (y) minus seven meters. A relationship between x and y can be derived using the area formula, leading to the equation: (1.5x) * x = (3y - 7) * y. When substituting y with x + 1, the values of x and y can be calculated. The discussion emphasizes the importance of showing attempted work for better assistance. Understanding the relationships and calculations is crucial for solving the problem effectively.
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Two rectangular tables are equal in area. The length of the first plot is on and a half times its width. The length of the second plot is seven (7) metre less than three times its width.

a) Denoting the width of the first plot by x meters and the width of the second plot by y meters, derive a relationship between x and y.

b) If y=x+1, calculate the values of x an y
 
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Can you show us what you have tried? Our helpers are better able to help if we can see exactly where you are stuck and what you have done. :D
 
MarkFL said:
Can you show us what you have tried? Our helpers are better able to help if we can see exactly where you are stuck and what you have done. :D

thats the problem i really don't understand it i really don't know where to start mathematics is a little difficult for me at times
 
Okay, let's look at what we are given:

Two rectangular tables are equal in area. The length of the first plot is one and a half times its width. The length of the second plot is seven (7) meters less than three times its width.

a) Denoting the width of the first plot by x meters and the width of the second plot by y meters, derive a relationship between x and y.

b) If y=x+1, calculate the values of x and y

For a rectangle, we know:

Area = Width times Length

For the first plot we are told:

The length of the first plot is one and a half times its width.

We are told to denote the width of the first plot by $x$. If the length is one and a half times the width, then how may we express the length of this first plot in terms of the width $x$?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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