the the length of a rectangle is 2 cm less than three times its width. its area is 21 cm. find the dimensions of rectangle.

The width is ? ,

the length is ?

thanks

the the length of a rectangle is 2 cm less than three times its width. its area is 21 cm. find the dimensions of rectangle.

The width is ? ,

the length is ?

thanks

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Let W represent the width of the rectangle and L represent the length of the rectangle. That is,

width = W and length = L

We are given that the length (L) is 2 cm less than 3 times the width (W). From this, we arrive at the following:

L = 3W - 2

We are also given that the area (A) of the rectangle is 21 cm^{2}. Since the area of a rectangle is the product of the rectangle's width and the length, then we have the following:

area = A = 21 cm^{2}

A = W•L ==> 21 = W•(3W - 2)

Distributing the W into each term inside the parentheses we get the following:

21 = W•3W - W•2

21 = 3W^{2} - 2W

Subtracting 21 from both sides of the equation yields the following quadratic equation:

3W^{2} - 2W - 21 = 0

Solve by factoring the left hand side of the above equation:

(3W + 7)(W - 3) = 0

By the zero product property we can set each binomial equal to 0 and solve for W in each:

3W + 7 = 0 and W - 3 = 0

3W = -7 W = 3

W = -7/3

Since the width cannot be negative, we omit the negative answer and only consider the positive answer. That is, the width (W) of the rectangle is equal to 3 cm.

Since the length (L) of the rectangle is 2 cm less than 3 times the width, then the length of determined by using the expression we found for L:

L = 3W - 2 = 3(3) - 2 = 9 - 2 = 7

Thus, the length of the rectangle is 7 cm.

Hey Motammem -- sometimes an "eyeball" approach works:

Begin with focus on the simplest part of the problem ... Area is 21 ... probably 3x7

Finally check 3(3) is 2 bigger than 7 ... Very best wishes :)