Two times the least of three consecutive odd integers exceeds three times the greatest by 15. What are the integers?

Let x be the least of the three consecutive old integers.
Since consecutive old integers are 2 away from each other, the next two are $x+2$ and $x+4$.
Two times the least is then 2x and three times the gretaest is then $3(x+4)$.
2x exceeds $3(x+4)$ by 15 which means the difference of 2x and $3(x+4)$ is equal to 15.
$2x-3(x+4)=15$
Distribute the -3 to x and 4.
$2x-3(x)-3(4)=15$
$2x-3x-12=15$
Combine the like terms of 2x and -3x on the left side.
$-x-12=15$
Add 12 on both sides.
$-x-12+12=15+12=x=27$
Multiply both sides by -1.
$-x\ast -1=27\ast -1$
$x=-27$
Find the order two old integers by substituing in $x=-27$ into $x+2$ and $x+4$.
$x+2=-27+2=-25$
$x+4=-27+4=-23$
Results: -27, -25, and -23