Question 164470
Let {{{a}}}= youngest age
Let {{{b}}}= middle one's age
Let {{{c}}}= oldest one's age
The youngest would split half of her silver dollars evenly with the other two sisters.
The middle and oldest sisters get {{{a/4}}} and {{{a/4}}} silver dollars
The middle, who now has {{{b + a/4}}} gives {{{4}}} to the others
So, the middle now has {{{b + a/4 - 8}}}
The youngest has {{{a/2 + 4}}}
The oldest has {{{c + a/4 + 4}}}
Now the oldest gives 1/2 her money to the other sisters
The oldest now has {{{c/2 + a/8 + 2}}}
Each sister gets {{{(1/4)*(c + a/4 + 4)}}}
So, the middle one has {{{c/4 + a/16 + 1 + b + a/4 - 8}}}
And, the youngest has {{{c/4 + a/16 + 1 + a/2 + 4}}}
Now each girl has 12 silver dollars
{{{c/4 + a/16 + 1 + a/2 + 4 = 12}}}
{{{9a/16 + c/4 + 5 = 12}}} (youngest)
{{{5a/16 + b + c/4 - 7 = 12}}} middle
{{{2a/16 + c/2 + 2 = 12}}} oldest
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Multiply each equation by {{{16}}}
(1) {{{9a + 4c + 80 = 12*16}}}
(2) {{{5a + 16b + 4c - 112 = 12*16}}}
(3) {{{2a + 8c + 32 = 12*16}}}
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Multiply (1) by {{{2}}}, then subtract (3) from (1)
(1) {{{18a + 8c + 160 = 2*12*16}}}
(3) {{{-2a - 8c -32 = -12*16}}}
{{{16a + 128 = 12*16}}}
{{{16a = 192 - 128}}}
{{{16a = 64}}}
{{{a = 4}}}
Substitute this in (3)
(3) {{{2*4 + 8c + 32 = 192}}}
{{{8c = 192 - 40}}}
{{{8c = 152}}}
{{{c = 19}}}
Substitute {{{a}}} and {{{c}}} in (2)
(2) {{{5*4 + 16b + 4*19 - 112 = 192}}}
{{{20 + 16b + 76 - 112 = 192}}}
{{{16b = 192 + 112 - 20 - 76}}}
{{{16b = 208}}}
{{{b = 13}}}    
Their ages are 4,13 and 19