Which equation can be used to find two numbers such that twice the first equals three times the second and three times their difference exceeds twice their difference by 13?

Let the numbers be a and b and b>a

2a=3b and 3 (b-a) = 2 (b-a) + 13

Solving the first for a:

2a=3b

a=3b/2, now using this value of a in the second equation gives you:

3 (b-3b/2) = 2 (b-3b/2) + 13 upon performing indicated operations.

3b-9b/2=2b-6b/2+13 making all terms have a common denominator of 2

(6b-9b) / 2 = (4b-6b+26) / 2 multiplying the whole equation by 2

6b-9b=4b-6b+26 combining like terms

-3b=-2b+26 adding 2b to both sides

-b=26 dividing both sides by - 1

b=-26, since a=3b/2

a=3 (-26) / 2

a=-39

So the numbers are - 39 and - 26

check ...

2a=3b becomes:

2 (-39) = 3 (-26)

-78=-78

3 (b-a) = 2 (b-a) + 13 becomes:

3 (-26--39) = 2 (-26--39) + 13

3 (13) = 2 (13) + 13

39=26+13

39=39

correct : P