SOLUTION: Two rectangles A and B each have an area of 11 cm^2 The length of rectangle A is x cm The length of rectangle B is (x+3) cm Given that the width of rectangle A is 2 cm greater t

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Question 1145118: Two rectangles A and B each have an area of 11 cm^2
The length of rectangle A is x cm
The length of rectangle B is (x+3) cm
Given that the width of rectangle A is 2 cm greater than the width of rectangle B form an equation in x and show that it simplifies to
2x^2 + 6x - 33 = 0
Found 3 solutions by josmiceli, ikleyn, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width of B = +y+
The width of A = +y+%2B+2+
-----------------------------------
(1) +y%2Ax+=+11+
(2) +%28+y+%2B+2+%29%2A%28+x+%2B+3+%29+=+11+
------------------------------------
(1) +y+=+11%2Fx+
and
{2} +y+%2B+2+=+11%2F%28+x+%2B+3+%29+
(2) +y+=+11%2F%28+x+%2B+3+%29+-+2+
------------------------------------
Set (1) = (2)
+11%2Fx+=+11%2F%28+x+%2B+3+%29+-+2+
Multiply both sides by +x%2A%28+x+%2B+3+%29+
+11%2A%28+x+%2B+3+%29+=+11x+-+2x%2A%28+x+%2B+3+%29+
+11x+%2B+33+=+11x+-+2x%5E2+-+6x+
+33+=+-2x%5E2+-+6x+
+2x%5E2+%2B+6x+%2B+33+=+0+
--------------------------------
The sign of the constant term is different. Check my math.

Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The solution by @josmiceli is INCORRECT. 

              His equation (2) is incorrect, leading to wrong answer.

            I came to bring the correct solution.


Solution 

From the condition, the width of rectangle  A  is  11%2Fx cm.

                    the width of rectangle  B  is 11%2F%28x%2B3%29 cm.


Also, from the condition


    11%2Fx - 11%2F%28x%2B3%29 = 2  cm.      (1)


It is your basic equation, and at this point, the setup is just completed.


Now my (and your) major task is to transform this equation to the standard form of a quadratic equation.


For it, multiply both side of the equation (1) by  x*(x+3).  You will get


    11*(x+3) - 11x = 2x*(x+3)

    11x + 33 - 11x = 2x^2 + 6x

    2x^2 + 6x - 33 = 0.


Your question is answered:  The equation in your post IS CORRECT.      ANSWER


Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Two rectangles A and B each have an area of 11 cm^2
The length of rectangle A is x cm
The length of rectangle B is (x+3) cm
Given that the width of rectangle A is 2 cm greater than the width of rectangle B form an equation in x and show that it simplifies to
2x^2 + 6x - 33 = 0
With the length and area of rectangle A being x and 11, respectively, the width of rectangle A becomes: 11%2Fx

With the length and area of rectangle B being x + 3 and 11, respectively, the width of rectangle B becomes: 11%2F%28x+%2B+3%29

As width of rectangle A is 2 cm GREATER than width of rectangle B, we get: matrix%281%2C3%2C+11%2Fx%2C+%22=%22%2C+11%2F%28x+%2B+3%29+%2B+2%29

11(x + 3) = 11x + 2x(x + 3) ------- Multiplying by LCD, x(x + 3)