Question

A square and a rectangle have the same area. The length of the rectangle is three inches more than the side of the square. The width of the rectangle is two inches less than the side of the square. Algebraically, find the dimensions of both the square and the rectangle.

Answer 1

The side length of the square is
`6~\text{inches}`.

The length of the rectangle is
`9~\text{inches}` , and the width is
`4~\text{inches}`.

Explanation:

Step 1: Assume your variables

Let's consider the area of both the square and the rectangle to be
`A`.

And let's consider the side of the square to be
`x`.

So, according to this variable, the length of the rectangle would be:
`x+3`, because it is three inches more than the square's side, which basically means 3 added to the side length, and thus
`x+3`.

Similarly, the width of the rectangle would be:
`x-2`

Step 2: Create your equations

The area of a square is given by:

`\text{Area}=\text{Side}^(2)`

We have presumed the side to be
`x`, and the area to be
`A`, so substitute those values into the formula:

`A=x^(2)`

The area of a rectangle is given by:

`\text{Area}=l* w`

We have found the length to be 
`x+3`, the width to be
`x-2`, and the area to be
`A`, so substitute these values into the formula:

`A=(x+3)(x-2)`

Step 3: Solve the equations

In the step above, we have found two different solutions for
`A`.

Since both the
`A` are equal to each other, we can write another equation:

`A=A`

And so, we can substitute the two different expressions for
`A` in this equation:

`A=A\\x^(2)=(x+3)(x-2)`

Simplify:

`x^(2)=(x+3)(x-2)\\\\\text{Use Distributive property:}\\x^(2)=x(x-2)+3(x-2)\\\\\text{Multiply:}\\x^(2)=x^(2)-2x+3x-6\\\\\text{Subtract}~x^(2)~\text{from both sides of the equation:}\\x^(2)-x^(2)=x^(2)-x^(2)-2x+3x-6\\\\\text{Simplify:}\\0=x-6\\\\\text{Add 6 to both sides of the equation:}\\0+6=x-6+6\\\\\text{Simplify:}\\6=x\\\\\text{Rearrange the equation:}\\x=6`

Step 3: Find the dimensions

Since
`x` represented the side length of the square, the side length of the square is
`6~\text{inches}`.

The length of the rectangle is
`9~\text{inches}` (
`6+3`), and the width is
`4~\text{inches}` (
`6-2`).

Answer 2

Areas of rectangle , and square. Comparison and find its lengths

Area of rectangle = width x length = W x L

Area of square = (side length )^ 2

Length = (side length) + 3

Width = (side length ) -2

Then

(width x length ) = (side length)^2

W x L = S^2

Now replace W and L

(S + 3 )• ( S - 2) = S^2

this is a second degree equation

evelopeD

Eliminate parenthesis

S^2 - 2S + 3S - 6 = S^2

eliminate S^2

-2S + 3S -6 = 0

S - 6 = 0

Then

S = 6. Its side length of square

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