Let the numbers be A and B.

The product of the two numbers is 24

=> A*B = 24

One number is 2 more than twice the other

=> A = 2 + 2*B

substitute this in A*B = 24

=> (2 + 2*B)*B = 24

=> 2B + 2B^2...

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Let the numbers be A and B.

The product of the two numbers is 24

=> A*B = 24

One number is 2 more than twice the other

=> A = 2 + 2*B

substitute this in A*B = 24

=> (2 + 2*B)*B = 24

=> 2B + 2B^2 = 24

=> B^2 + B - 12 = 0

=> B^2 + 4B - 3B - 12 = 0

=> B(B + 4) - 3(B + 4) = 0

=> (B - 3)(B + 4) = 0

=> B = 3 and B = 4

A = 8 and A = 6

**The numbers are (3, 8) and (4, 6)**

Let the numbers be x and y

Given that the product is 24.

==> x*y = 24.............(1)

Now we know that one number is 2 more than twice the other number.

==> x = 2y+ 2 ..............(2)

We will substitute (2) into (1).

==> (2y+2)*y = 24

==> 2y^2 + 2y = 24

We will divide by 2.

==> y^2 + y= 12

==> y^2 +y - 12 = 0

==> (y+4)(y-3) = 0

==> y1= -4 ==> x1= 2*-4+2 = -6

==> y2= 3 ==> x2= 2*3+2 = 8

Then we have two sets of solution.

**==> The numbers are 3 and 8 OR -4 and -6.**