consec. pos. integers w/ product of 1st, 3rd, less 2nd, is

A

acridcola

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A friend of mine is having a problem and I havent seen algebra in so long I cant really be of help Her problem is,

find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third

She wrote down

x
x+1
x+2

but after that shes having a hard time explaining it to me. I assume the answer is x = 5 or 5, 6, 7 and the equation is

x(x+2) -(x +1) = 4(x+2) + 1
x(x+2) -(x +1) = 29
4(x+2) = 28

am I wrong? and how do you write this as proof?
 
Re: consecutive positive integers

Seems to me ...

if the first is represented by x,
then the second is x+1 and
the third is x+2.

Now, build your equation by substituting into the phrase...

the product of the first and third, minus the second, is 1 more than 4 times the third

... remembering that "is" is "equals".

Good luck.
 
Re: consecutive positive integers

acridcola said:
A friend of mine is having a problem and I havent seen algebra in so long I cant really be of help

Her problem is,

find three consecutive positive integers such that the product of the first and third, minus the second, is 1 more than 4 times the third

She wrote down

x
x+1
x+2

but after that shes having a hard time explaining it to me.

I assume the answer is x = 5 or 5, 6, 7 and the equation is

x(x+2) -(x +1) = 4(x+2) + 1

x(x+2) -(x +1) = 29
4(x+2) = 28 <<<<< Correct

am I wrong? and how do you write this as proof?

There is no proof to it other than retracing your path

the product of the first and third, = 7*5 = 35

minus the second = 35-6 = 29

4 times the third = 4*7 = 28

1 more than = 28 + 1 = 29

and it checks...
Click to expand...
 
Re: consecutive positive integers

A bit easier if you let the 3 numbers be: x-1, x and x+1

(x-1)(x+1) - x = 4(x+1) + 1
x^2 - 1 - x = 4x + 4 + 1
x^2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6 or x = -1 : so x = 6
 
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