Question 5 the area of a rectangle is 35 ft2 , and the length of the rectangle is 3 ft more than twice the width. find the dimensions of the rectangle.

Let's assume w represent the width of the rectangle.

Given that, the length of the rectangle is 3 ft more than twice the width.

So, we can write:

Length = 3 + 2w

And area = 35

So, length * width = 35

(3+2w) * w = 35

3w +2w²=35 By distribution property.

3w + 2w²-35 = 0 Subtract 35 from each sides.

2w² + 3w - 35 = 0

Next step is to solve the above equation by factoring to get the vaue of w.

To factor the above trinomial, multiply the constant -35 with the leading coefficient 2.

So, -35*2 = -70.

Next step is to breakdown -70 into two multiples so that their addition will result the coefficient of w=3.

So, -70 = -7 * 10

And addition of -7 and 10 will give 3.

So, next step is to replace 3w with -7w + 10w. Therefore,

2w² - 7w + 10w - 35 = 0

(2w² - 7w) + (10w - 35) = 0 Make the group of two terms.

w (2w - 7) + 5 (2w - 7) = 0 Take out the common factor from each group.

(2w - 7) ( w + 5) = 0 Again take out the common factor (2w-7).

So, 2w -7 =0 and w + 5 =0 Set both the factor equal to 0.

2w = 7 and w = -5

w = 7/2

So, w = 3.5 because width cannot be negative 5.

Now plug in w=3.5 in 3 + 2w to get the length.

So, length = 3 + 2(3.5)

= 3 + 7

= 10

Hence the dimension of the rectangle is 10 ft and 3.5 ft.

b

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she gotta workk that a 2

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