The length of a rectangle is 4 units more than its width. The area of the rectangle is 25 more than 4 times the width. What is the width of the rectangle?

The wording is poor.

Did you mean:
" The numerical value of the area of the rectangle is 25 more than the numerical value of 4 times the width" ?

if so, then ...

width --- x
length -- x+4

x(x+4) = 4x + 25
x^2 + 4x = 4x + 25
x^2 = 25
x = 5

The width is 5 units
(length is 9 units)

check:
area = 45
is 45 greater than 4(5) by 25 ?
yes it is

The length of rectangle is 2x+5 units and the breadth is 3x_1 units .find the perimeter and area of the rectangle

To find the width of the rectangle, let's set up an equation based on the given information.

Let's assume that the width of the rectangle is "w" units. According to the given information, the length of the rectangle is 4 units more than its width, so the length can be expressed as "w + 4".

The area of a rectangle is given by the formula: Area = Length x Width. In this case, the area is 25 more than 4 times the width, so we can write the equation as:

(w + 4) * w = 4w + 25

Now, let's solve the equation to find the value of "w" (the width):

w^2 + 4w = 4w + 25

Simplifying the equation, we get:

w^2 + 4w - 4w - 25 = 0

w^2 - 25 = 0

Now, we can factorize the equation:

(w - 5)(w + 5) = 0

From the factored form, we have two possible solutions for w:

w - 5 = 0 --> w = 5

w + 5 = 0 --> w = -5 (not a valid solution as width cannot be negative)

Therefore, the width of the rectangle is 5 units.