Question 280051
Let {{{g}}} = original number of girls
Let {{{b}}} = original number of boys
given:
Initially, there are:
{{{b + g}}} in the group
20 girls leave
{{{b + (g - 20)}}} in the group
(1) {{{b = 2*(g - 20)}}}
Then {{{60}}} boys leave
{{{(b - 60) + (g - 20)}}} in the group
(2) {{{g - 20 = 2*(b - 60)}}}
(1) and (2) are 2 equations with 2 unknowns, so
it's solvable
(1) {{{b = 2*(g - 20)}}}
(1) {{{b = 2g - 40}}}
(1) {{{2g - b = 40}}}
(2) {{{g - 20 = 2*(b - 60)}}}
(2) {{{g - 20 = 2b - 120}}}
(2) {{{-g + 2b = 100}}}
Multiply both sides of (1) by {{{2}}} and add equations
(1) {{{4g - 2b = 80}}}
(2) {{{-g + 2b = 100}}}
{{{3g = 180}}}
{{{g = 60}}}
The number of girls in the original group was 60
check answer:
(1) {{{b = 2*(g - 20)}}}
{{{b = 2*(60 - 20)}}}
{{{b = 80}}}
(2) {{{g - 20 = 2*(b - 60)}}}
(2) {{{60 - 20 = 2*(80 - 60)}}}
{{{40 = 2*20}}}
{{{40 = 40}}}
OK