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Mathematics 17 Online
OpenStudy (anonymous):

The length of a rectangle is 3 inches less than twice the width. The area is 65 square inches. Find the length and width

OpenStudy (anonymous):

How would you set this problem up using x as a variable for the width?

OpenStudy (anonymous):

I have no idea

OpenStudy (anonymous):

Your first sentence gives you all you need: the length is 3 less (subtraction) than twice (multiplication) the width. So if you multiply the width by two (2x), then what would you do?

OpenStudy (anonymous):

I'm not sure

OpenStudy (anonymous):

x=13/2

OpenStudy (anonymous):

width = x length = 2x - ? Area = length * width

OpenStudy (anonymous):

3?

OpenStudy (anonymous):

Yup! length = 2x - 3 Now, you plug in these values for length and width into the equation for area: Area = length * width Area = (2x - 3) (x) distribute the x and what do you get?

OpenStudy (anonymous):

How do I distribute?

OpenStudy (anonymous):

(x*2x) - (x*3)

OpenStudy (anonymous):

3x-3x?

OpenStudy (anonymous):

Close, \[2x^2-3x\] because \[x \times x=x^2\]

OpenStudy (anonymous):

Now plug in your given value for Area (65): \[65=2x^2-3x\] and solve for x

OpenStudy (anonymous):

How do I solve that?

OpenStudy (anonymous):

2x^2-3x+65=0?

OpenStudy (anonymous):

Almost, subtract 65 from both sides to get zero on one side \[0=2x^2-3x-65\]

OpenStudy (anonymous):

65-65=0

OpenStudy (anonymous):

Okay. But what do I do now?

OpenStudy (anonymous):

Kinda hard to explain, but here http://www.purplemath.com/modules/solvquad4.htm

OpenStudy (anonymous):

Thank you

OpenStudy (anonymous):

No problem! :)

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