Answer :
Firstly, let x represent Alex's age, y represent George's age and z represent Carl's age.
from the question;
Alex is 12 years older than George, So;
[tex]x=y+12\ldots\ldots\ldots\ldots.1[/tex]Carl is three times older than Alex, So;
[tex]z=3x\ldots\ldots\ldots..2[/tex]The sum of their ages is 68, So;
[tex]x+y+z=68\ldots\ldots\ldots\ldots\ldots3[/tex]Now we have three equations and three unknowns, so it is solvable.
Let us substitute equation 2 into equation 3; that is replace z with 3x in equation 3.
[tex]\begin{gathered} x+y+3x=68 \\ 4x+y=68\ldots\ldots\ldots\ldots\ldots\ldots4 \end{gathered}[/tex]Next, let us substitute equation 1 into equation 4. that is replace x with y+12 in equation 4.
[tex]\begin{gathered} 4(y+12)+y=68 \\ 4y+48+y=68 \\ 5y+48=68\ldots\ldots\ldots.5 \end{gathered}[/tex]we can now solve for the value of y from equation 5.
[tex]\begin{gathered} 5y+48=68\ldots\ldots\ldots.5 \\ \text{subtract 48 from both sides.} \\ 5y+48-48=68-48 \\ 5y=20 \\ y=\frac{20}{5} \\ y=4 \end{gathered}[/tex]let us now replace y with 4 in equation 1 to get the value of x. since y = 4;
[tex]\begin{gathered} x=y+12\ldots\ldots\ldots\ldots.1 \\ x=4+12 \\ x=16 \end{gathered}[/tex]then, since x =16 let us replace x with 16 in equation 2 to get z.
[tex]\begin{gathered} z=3x\ldots\ldots\ldots..2 \\ z=3(16) \\ z=\text{ 48} \end{gathered}[/tex]so we have;
[tex]\begin{gathered} \text{Alex's age = x = 4 years} \\ George^{\prime}sage_{}=y=16\text{ years} \\ Carl^{\prime}sage=z=48\text{ years} \\ \end{gathered}[/tex]We now need to find the ratio of George, Carl and Alex's age.
[tex]\begin{gathered} 16\colon48\colon4 \\ \text{dividing through by 4 we have;} \\ 4\colon12\colon1 \end{gathered}[/tex]So the ratio of their ages are;
[tex]4\colon12\colon1[/tex]