Question 183442
Let {{{m}}}= Maya's age now
Let {{{d}}}= David's age now
In 15 years, Maya's age = {{{m + 15}}}
In 15 years, David'd age = {{{d + 15}}}
In 10 years, Maya's age = {{{m + 10}}}
given:
(1) {{{m + 15 = 2d}}}
(2) {{{d + 15 = m + 10}}}
This is 2 equations and 2 unknowns, so it's solvable
(1) {{{m + 15 = 2d}}}
(2) {{{m + 10 = d + 15}}}
-------------------------
(1) {{{m - 2d + 15 = 0}}}
(2) {{{m - d - 5 = 0}}}
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Subtract (2) from (1)
-------------------------
(1) {{{m - 2d + 15 = 0}}}
(2) {{{-m + d + 5 = 0}}}
(3) {{{-d + 20 = 0}}}
(3) {{{d = 20}}}
Plug this result back into (2)
(2) {{{d + 15 = m + 10}}}
(2) {{{20 + 15 = m + 10}}}
(2) {{{35 = m + 10}}}
(2) {{{m = 25}}}
Maya's age now is 25 and David's age now is 20
check:
given:
(1) {{{m + 15 = 2d}}}
(1) {{{25 + 15 = 2*20}}}
(1) {{{40 = 40}}}
-----------------
(2) {{{d + 15 = m + 10}}}
(2) {{{20 + 15 = 25 + 10}}}
(2) {{{35 = 35}}}
OK