Posted by **J** on Thursday, September 17, 2009 at 10:09pm.

find three consecutive positive numbers such that the product of the first and third minus the second, is 1 more than 6 times the third

- algebra word problem -
**MathMate**, Thursday, September 17, 2009 at 10:17pm
" find three consecutive positive numbers"

Let the middle integer be M.

"such that the product of the first and third minus the second,"

(M-1)*(M+1) - M

"is 1 more than 6 times the third"

= 6(M+1)+1

The resulting formula is therefore

(M-1)(M+1)-M = 6(M+1)+1

This equation has rational factors, or you could use the quadratic formula to find the solution. The answer for M is less than 10.

- algebra word problem -
**J**, Thursday, September 17, 2009 at 10:40pm
I still don't understand it

- algebra word problem -
**DrBob222**, Thursday, September 17, 2009 at 10:51pm
This is worked exactly the same way as mathmate did it but uses a little different terminology. Perhaps this will help.

These are consecutive numbers; therefore, if we let x = first number, x + 1 = y is the second number and y+1 = z = the third number.

Then product of 1st and 3rd is

xz and that minus the second makes it

xz-y and that = 6(3rd)+1 but the 3rd is 6z so

xz-y=6z+1

Now just substitute for x = y-1 and for z = y+1 and solve for y (which is the middle number).

(y-1)(y-1) - y = 6(y+1)+1

y=??

y-1 = x

y+1 = z

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