Respuesta :

reinhard10158

Step-by-step explanation:

length = width + 5

58 = 2×length + 2×width

using the first in the second equation :

58 = 2×(width + 5) + 2×width = 2×width + 10 + 2×width =

= 4×width + 10

48 = 4×width

width = 12 cm

length = width + 5 = 12 + 5 = 17 cm

Given:

  • #1: The length of a rectangle of 5cm more than its width.
  • #2: Its (Rectangle) perimeter is 58cm

Correct #1: The length of a rectangle is 5 cm more than its width.

Correct #2: Its (Rectangle) perimeter is 58 cm

Determining an equation that represents (#1).

  • ⇒ #1: The length of a rectangle of 5 cm more than its width.

Let the length of the rectangle be represented as "L" and the width of the rectangle be represented as "W". Then, we get the following statement:

  • ⇒ #1: L is 5 cm more than its W.

The word "more" means "addition".

  • ⇒ #1: L is 5 cm + W

The word "is" defines that two terms or expressions are equivalent.

  • ⇒ #1: L = 5 cm + W
  • ⇒ #1: L = 5 + W

∴ We have successfully converted (#1) into an equation.

Determining an equation that represents (#2).

  • ⇒ #2: Its (Rectangle) perimeter is 58 cm

Let the perimeter of the rectangle be represented as "P". Then, we get the following statement:

  • ⇒ #2: P is 58 cm

The word "is" defines that two terms or expressions are equivalent.

  • ⇒ #2: P = 58 cm

∴ We have successfully converted (#2) into an equation.

Determining the length and the width:

Obtained equations:

  • L (Length) = 5 + W (Width)
  • P (Perimeter) = 58 cm

[tex]\boxed{\text{Perimeter of a rectangle: 2L + 2W \ \ \ \ (L = Length; W = Width)}}[/tex]

Therefore, we get the following equation:

[tex]\implies \text{2L + 2W = 58 cm}[/tex]

From the obtained equations, we can see that the measure of the length (L) is 5 + W, where the "W" is represented as the width of the rectangle.

[tex]\implies \text{2(5 + W) + 2W = 58 cm}[/tex]

Apply the distributive property rule: a(b + c) = ab + ac

[tex]\implies \text{10 + 2W + 2W = 58 cm}[/tex]

Combine like terms and simplify:

[tex]\implies \text{10 + 4W = 58 cm}[/tex]

Divide 4 on both sides to isolate the "W" variable from its coefficient.

[tex]\implies \dfrac{10 + \text{4W}}{4} = \dfrac{58 \ \text{cm}}{4}[/tex]

[tex]\implies \dfra{2.5 + \text{W} = \dfrac{58 \ \text{cm}}{4} = 14.5[/tex]

[tex]\implies \dfra{2.5 + \text{W} = 14.5[/tex]

Subtract 2.5 on both sides to isolate the "W" variable completely.

[tex]\implies \dfra{2.5 - 2.5 + \text{W} = 14.5 - 2.5[/tex]

[tex]\implies \text{W} = 14.5 - 2.5 = 12 \ \text{cm}[/tex]

Now, plug the width in the equation that represents the relationship between the length and the width, which is L = 5 + W.

[tex]\implies \text{L = 5 + W}[/tex]

[tex]\implies \text{L = 5 + 12}[/tex]

[tex]\implies \text{L = 17 cm}[/tex]

Therefore, the length and the width are 17 and 12 centimeters respectively.

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