We can write x=2n+1, where n is an integer. y=2n+3 and z=2n+5, because consecutive odd numbers are separated by 2.
3(2n+1)+2(2n+3)=4(2n+5)-5; 6n+3+4n+6=8n+20-5; 2n=6, n=3.
x=2n+1=7, y=9, z=11, assuming x is the first integer (and the smallest), y the second and z the third and the largest).
Other solutions, because the question doesn't relate x, y and z to the order of the numbers (e.g., is x the smallest or largest number? Is y the first integer? etc.):
3(2n+3)+2(2n+5)=4(2n+1)-5; 6n+9+4n+10=8n+4-5; 2n=-20, n=-10, so x=-17, y=-15, z=-19.
3(2n+5)+2(2n+1)=4(2n+3)-5; 2n=12-5-2-15=-10, n=-5, so x=-5, y=-9, z=-7.
3(2n+5)+2(2n+3)=4(2n+1)-5; 2n=4-5-15-6=-22, n=-11, so x=-17, y=-19, z=-21.
3(2n+1)+2(2n+5)=4(2n+3)-5; 2n=12-5-10-3=-6, n=-3, so x=-5, y=-1, z=-3.
3(2n+3)+2(2n+1)=4(2n+5)-5; 2n=20-5-9-2=4, n=2, so x=7, y=5, z=9.
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