SOLUTION: The dimensions of a rectangle are such that its length is 11in more than its width. If the length were doubled and if the width were decreased by 5 in, the area would be increase

Question 854124: The dimensions of a rectangle are such that its length is 11in more than its width.
If the length were doubled and if the width were decreased by 5 in, the area would be increased by 10in^2.
What are the length and width of the rectangle?
Found 3 solutions by aditya2117, MathTherapy, Vladdroid:
Answer by aditya2117(32)   (Show Source): You can put this solution on YOUR website!
Let the width be x inches.
Length = (x+11) inches
Area= x(x+11) sq.inches
Now As per question,
2(x+11)(x-5)= x(x+11) + 10^2
=> 2(x^2+6x-55)=x^2+11x+100
=> 2x^2+12x-110=x^2+11x+100
=> 2x^2-x^2+12x-11x-110-100=0
=> x^2 + x - 210 = 0
=> x^2 + 15x - 14x - 210 = 0
=> x(x+15)-14(x+15)=0
=> (x+15)(x-14)=0
Therefore,
x = -15 or 14
But measurement # negetive
So, x=14
x+11 = 25
Ans: Length is 25 inches and width is 14 inches.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

The dimensions of a rectangle are such that its length is 11in more than its width.
If the length were doubled and if the width were decreased by 5 in, the area would be increased by 10in^2.
What are the length and width of the rectangle?

Original width: inches
Original length: inches
You can do the check!!
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Answer by Vladdroid(91)   (Show Source): You can put this solution on YOUR website!
Let the width be x inches.
Length = (x+11) inches
Area= x(x+11) sq.inches
Now As per question,
2(x+11)(x-5)= x(x+11) + 10^2
=> 2(x^2+6x-55)=x^2+11x+100
=> 2x^2+12x-110=x^2+11x+100
=> 2x^2-x^2+12x-11x-110-100=0
=> x^2 + x - 210 = 0
=> x^2 + 15x - 14x - 210 = 0
=> x(x+15)-14(x+15)=0
=> (x+15)(x-14)=0
Therefore,
x = -15 or 14
But measurement # negetive
So, x=14
x+11 = 25
Ans: Length is 25 inches and width is 14 inches.
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