The dimensons of a rectangle are such that its length is 5 inches more than its width. If the lenght were doubled and if the width were decreased by 2 inches, the area would be increased by 112 inches squared. What are the length and width of the rectangle?

Question

anonymous · Answer

Hi Welcome to Openstudy!

You are given:
width= x
length=5 + x  (length is 5 inches more than its width)

Then it says:
 If the lenght were doubled : 2(5 + x)
the width were decreased by 2 inches: (x-2)
therefore you'll have:
length: 2(5+x-2)=2(x+3) = 2x + 6
width= x-2

Then area would be increased by 112 inches squared.
Area of a rectangle is $A=l \times w$

so next we have to do is combine these formulas or simplify them..
do you understand so far? 

Original Area without conditions is: 
$A= (x+5)(x)$

New Area with the condition is:
$A+112= (x+3)(x-2)$

anonymous · Answer

*I mean combine and simplify $A=(x+5)(x)\ and\ A+112=(x+3)(x-2)$

anonymous · Answer

@Missy1101 still there?

anonymous · Answer

so let's simplify the equations:
$A=(x+5)(x)=x^2+5x$

$A+112=(x+3)(x-2)=x^2+x-6\A=x^2+x-6-112\A=x^2+x-118$

combine these two equations:
$x^2+5x = x^2+x-118$ since A is common between these equations..

Now it's your turn to continue..
can you simplify: $x^2+5x = x^2+x-118$ ? and calculate for the value of x (width) 
I'll wait for your response :)