Greetings!
To solve this problem, we will have to create a linear system:
Let Statements:
Let x represent the width of the first rectangle
Let y represent the length of the first rectangle
Linear System:
We can create the two equations (a system) using the information from the problem:
\(\left \{ {{x+2=y} \atop {(x-3)(y)=24}} \right.\)
Solve the system using Elimination or Substitution.
Isolate for x, in Equation #1:
\(x+2=y\)
\(x=y-2\)
Substitute this value into Equation #2:
\((x-3)(y)=24\)
\(((y-2)-3)(y)=24\)
Simplify the Equation:
\((y-2-3)(y)=24\)
\((y-5)(y)=24\)
Distribute the Parenthesis:
\((((y)(y))-(5)(y))=24\)
\(y^2-5y=24\)
Add -24 to both sides:
\((y^2-5y)+(-24)=(24)+(-24)\)
\(y^2-5y-24=0\)
Factor the Simple Trinomial:
\(y^2-8y+3y-24=0\)
\(y(y-8)+3(y-8)=0\)
\((y-8)(y+3)=0\)
Set Factors to equal 0:
First Factor:
\(y-8=0\)
\(y=8\)
Second Factor:
\(y+3=0\)
\(y=-3\)
Since it is impossible to have a "negative length" the only possible answer would be:
\(\boxed{y=8}\)
Using this value, find the value of x:
\(x+2=y\)
\(x+2=(8)\)
Add -2 to both sides:
\((x+2)+(-2)=(8)+(-2)\)
\(\boxed{x=6}\)
To find the area of the first rectangle, we can use a formula:
\(A_{Rectangle}=(l)(w)\)
Input the values:
\(A_{Rectangle}=(8)(6)\)
Simplify:
\(A_{Rectangle}=48\)
The Area of the First Rectangle is:
\(\boxed{=48in^2}\)
I hope this helped!
-Benjamin
