Question

The perimeter of a rectangle is 40cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm. Find the dimensions of the original rectangle.
Create a linear system to model the situation.
Determine, by substitution, the dimensions of the original rectangle (sketch and label rectangles to help) PLS SHOW ALL WORK NEED ASAP

Answer 1

Answer:

The dimensions of the original rectangle are Length =12cm, Width=8cm

Step-by-step explanation:

Let the length of the rectangle be x

Let the width of the rectangle be y

The perimeter of a rectangle is 40cm.

2(x+y)=40

Divide both sides by 2

x+y=20

If the length were doubled(2x) and the width halved(y/2), the perimeter would be increased by 16cm,i.e.(40+16)cm

Therefore:

$2(2x+\dfrac{y}{2})=56$

Divide both sides by 2

$2x+\dfrac{y}{2}=28$

From the first equation, x=20-y.

Substitute x=20-y into $2x+\dfrac{y}{2}=28$

$2x+\dfrac{y}{2}=28$

$2(20-y)+\dfrac{y}{2}=28\\\dfrac{4(20-y)+y}{2}=28\\80-4y+y=56\\-3y=-24\\ Divide both sides by -3\\y=8cm\\Recall:x=20-y\\x=20-8=12cm\\The dimensions of the original rectangle are Length =12cm, Width=8cm$