Question 396654
Let {{a}}} = Abbey's age now
Let {{{c}}} = Charles age now
Let {{{m}}} = Monte's age now
{{{a - 10}}} = Abbey's age 10 years ago
{{{c -10}}} = Charles age 10 years ago
{{{m - 3}}} = Monte's age 3 years ago
{{{c - 3}}} = Charles age 3 years ago
{{{a + 15}}} = Abbey's age in 15 years
{{{m + 15}}} = Mote's age in 15 years
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given:
(1) {{{a - 10 = 2*(c - 10)}}}
(2) {{{4*(m - 3) = 3*(c - 3)}}}
(3) {{{a + 15 = m + 15 + 8}}}
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This is 3 equations and 3 unknowns, so it's solvable
Rewriting:
(1) {{{a - 10 = 2c - 20}}}
(1) {{{a - 2c = -10}}}
and
(2) {{{4m - 12 = 3c - 9}}}
(2) {{{4m - 3c = 3}}}
and
(3) {{{a = m + 8}}}
(3) {{{a - m = 8}}}
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Multiply both sides of (3) by {{{4}}} and 
add (2) and (3)
(2) {{{4m - 3c = 3}}}
(3) {{{-4m + 4a = 32}}}
{{{4a - 3c = 35}}}
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Now multiply both sides of (1) by {{{4}}}
and subtract (1) from last result
{{{4a - 3c = 35}}}
 {{{-4a + 8c = 40}}}
{{{5c = 75}}}
{{{c = 15}}}
and, from (1):
(1) {{{a - 2*15 = -10}}}
{{{a - 30 = -10}}}
{{{a = 20}}}
and, from (3):
(3) {{{a - m = 8}}}
{{{20 - m = 8}}}
{{{m = 12}}}
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20 is Abbey's age now
15 is Charles age now
12 is Monte's age now
check answers:
(1) {{{20 - 10 = 2*(15 - 10)}}}
(1) {{{10 = 2*5}}}
{{{10 = 10}}}
and
(2) {{{4*(12 - 3) = 3*(15 - 3)}}}
(2) {{{4*9 = 3*12}}}
{{{36 = 36}}}
and
(3) {{{20 + 15 = 12 + 15 + 8}}}
(3) {{{35 = 35}}}
OK