The length of a rectangle is 3 inches less than twice the width. The area is 65 square inches. Find the length and width
How would you set this problem up using x as a variable for the width?
I have no idea
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?
I'm not sure
x=13/2
width = x length = 2x - ? Area = length * width
3?
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?
How do I distribute?
(x*2x) - (x*3)
3x-3x?
Close, \[2x^2-3x\] because \[x \times x=x^2\]
Now plug in your given value for Area (65): \[65=2x^2-3x\] and solve for x
How do I solve that?
2x^2-3x+65=0?
Almost, subtract 65 from both sides to get zero on one side \[0=2x^2-3x-65\]
65-65=0
Okay. But what do I do now?
Kinda hard to explain, but here http://www.purplemath.com/modules/solvquad4.htm
Thank you
No problem! :)
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