Question 12331
youngest=x
middle=y
oldest=z


The youngest kept half and gave 1/4 each to her sisters
x:{{{x/2}}}
y:Y+{{{x/4}}}
z:Z+{{{x/4}}}


The middle gave the oldest and youngest 4 each
x:{{{x/2}}}+4
y:Y+{{{x/4}}}-8
z:Z+{{{x/4}}}+4



the oldest kept half z so (Z+{{{x/4}}}+4) becomes:{{{z/2}}}+{{{x/8}}}+2
she gives half of that to each sister so they receive:  ({{{z/4}}}+{{{x/16}}}+1)

x:{{{x/2}}}+4+({{{z/4}}}+{{{x/16}}}+1)=16
y:Y+{{{x/4}}}-8+({{{z/4}}}+{{{x/16}}}+1)=16
z:{{{z/2}}}+{{{x/8}}}+2=16



{{{x/2}}}+4+({{{z/4}}}+{{{x/16}}}+1)=16   <--subtract {{{x/2}}} and 4 from both sides
({{{z/4}}}+{{{x/16}}}+1)=16-4-{{{x/2}}}
({{{z/4}}}+{{{x/16}}}+1)=12-{{{x/2}}}


y+{{{x/4}}}-8+({{{z/4}}}+{{{x/16}}}+1)=16  <--subtract y and {{{x/4}}} and add 8 to both sides 
({{{z/4}}}+{{{x/16}}}+1)=16+8-y-{{{x/4}}}
({{{z/4}}}+{{{x/16}}}+1)=24-y-{{{x/4}}}  



since 12-{{{x/2}}} and 24-y-{{{x/4}}}  both equal ({{{z/4}}}+{{{x/16}}}+1), you can equate them and solve for y:
12-{{{x/2}}}=24-y-{{{x/4}}}
12-{{{2x/4}}}=24-y-{{{x/4}}}   <--add {{{2x/4}}} to both sides
12=24-y+{{{x/4}}}     <--add y to both sides
12+y=24+{{{x/4}}}      <--subtract 12 from both sides
y=12+{{{x/4}}}


to solve for z: 
{{{z/2}}}+{{{x/8}}}+2=16  <--subtract 2 from both sides
{{{z/2}}}+{{{x/8}}}=14     <--subtract {{{x/8}}} from both sides
{{{z/2}}}=14-{{{x/8}}}    <--multiply both sides by 2
z=28-{{{x/4}}}



plug it back to x+y+z=48
x+(12+{{{x/4}}})+(28-{{{x/4}}})=48
x+{{{x/4}}}-{{{x/4}}}+12+28=48
x+40=48
x=8



since the youngest=x, she is 8 years old
middle y=(12+{{{x/4}}})=12+{{{8/4}}}=12+2=14
oldest z=28-{{{x/4}}}=28-{{{8/4}}}=28-2=26 



They are 8, 14, and 26 years old.