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April 24, 2014

April 24, 2014

Posted by **Samantha** on Saturday, March 14, 2009 at 2:18pm.

- college math -
**bobpursley**, Saturday, March 14, 2009 at 2:20pmN1=8+2*N2

N1*N2=8+2(N1+N2)

- college math -
**drwls**, Saturday, March 14, 2009 at 2:33pmLet the smaller number be x; that means the other number must be 2x + 8.

Solve the equation

x(2x+8) = 2x + 8 + x + 8

which can also be written

2x^2 + 8x = 3x + 16

or

2x^2 + 5x -16 = 0

There seems to be no integer solution. The smaller number is

x = (-5 + sqrt 153)/4 = 1.84233

and the larger number is 2x + 1 = 11.68466

Their sum is 13.5267

and the product is 21.527

as required.

- college math -
**GanonTEK**, Sunday, March 15, 2009 at 8:22amdrwls, you made a slight mistake - you forgot a factor of 2 in the RHS of the equation.

If N1 = x and N2 = 2x + 8

then the equation is

x(2x+8) = 8 + 2*(2x + 8 + x)

Which simplifies to

x^2 - x - 12 = 0

which has two integer solutions.

Hope that helps!

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