Solve the following word problem.

The length and width of a rectangle are (x + 4) cm and x cm. Respectively, a square has the same perimeter as this rectangle. If you add the areas of the rectangle and the square, the result is 94cm^2. Solve for x.

Note: Write your answer as x = ____ and x = ____

If the square has side s, then

4s = 2(x+4 + x)
s^2 + x(x+4) = 94

...

Area rectangle = x (x + 4) = x^2 + 4x

Perimeter of rectangle = 2 (x + x + 4)= 4x + 8
Perimeter of square = perimeter of rectangle = 4x + 8
Therefore length of square side = .25 (4x + 8) = x + 2
and area of square = (x + 2)^2 = x^2 + 4x + 4
Area of square + area of rectangle = 94 cm^2
Therefore:
x^2 + 4x + x^2 + 4x + 4 = 94
Or, after some rearrangement of terms:
x^2 + 4x - 45 = 0
Use quadratic theorum to obtain:
x = 5 cm or -9 cm
Choose x = 5 cm since a negative area is not physically real.
QED

To solve this word problem, we need to set up equations based on the given information and then solve for x.

Step 1: Let's start by finding the perimeter of the rectangle and the square.

Perimeter of a rectangle = 2(length + width)
Perimeter of the rectangle = 2((x + 4) + x) = 2(2x + 4) = 4x + 8

Perimeter of a square = 4(side length)
Since the square has the same perimeter as the rectangle, its perimeter = 4x

Step 2: Since the rectangle and square have the same perimeter, we can set up an equation using the perimeters.

4x + 8 = 4x

Simplifying the equation, we can subtract 4x from both sides:
8 = 0

However, we see that we have ended up with an incorrect statement, implying that there is no solution to this problem. This means that the given problem is inconsistent or has no valid answer. It could be that there was an error in the problem statement or some missing information.

Hence, we cannot solve for x in this word problem.