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

[ladybutterz]

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

 

[MarkFL]

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

 

[ladybutterz]

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

 

[MarkFL]

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 $x$?

 

[CellCoree]

a) The area of a rectangle is calculated by multiplying its length by its width. Since both rectangular tables are equal in area, we can set up the following equation:

Area of first plot = Area of second plot

Length of first plot * Width of first plot = Length of second plot * Width of second plot

Since the length of the first plot is one and a half times its width, we can represent it as 1.5x, where x is the width. Similarly, the length of the second plot can be represented as 3y-7, where y is the width. Substituting these values into our equation, we get:

1.5x * x = (3y-7) * y

1.5x^2 = 3y^2 - 7y

b) Since y=x+1, we can substitute this into our equation:

1.5x^2 = 3(x+1)^2 - 7(x+1)

1.5x^2 = 3x^2 + 6x + 3 - 7x - 7

1.5x^2 = 3x^2 - x - 4

0 = 1.5x^2 - 3x^2 + x + 4

0 = -1.5x^2 + x + 4

Using the quadratic formula, we can solve for the values of x:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in our values, we get:

x = (-1 ± √(1 - 4(-1.5)(4))) / 2(-1.5)

x = (-1 ± √(25)) / -3

x = (-1 ± 5) / -3

x = 4 or -0.5

Since we cannot have a negative width, the value of x is 4 meters. Therefore, the width of the first plot is 4 meters and the width of the second plot is 5 meters (since y=x+1).