Question 339215
Let number of children attending = {{{c}}}
Let number of students attending = {{{s}}}
Let number of adults attending = {{{a}}}
given:
Receipts from children = {{{4c}}}
Receipts from students = {{{6s}}}
Receipts from adults = {{{8a}}}
(1) {{{4c + 6s + 8a = 5600}}}
There was full attendance, so
(2) {{{c + s + a = 900}}}
(3) {{{a = (1/2)*(c + s)}}}
--------------------------
There are 3 equations and 3 unknowns, so
it should be solvable
From (3):
(3) {{{a = (1/2)*(c + s)}}}
{{{c + s = 2a}}}
Now substitute into (2)
(2) {{{2a + a = 900}}}
{{{3a = 900}}}
{{{a = 300}}}
From (1):
(1) {{{4c + 6s + 8*300 = 5600}}}
{{{4c + 6s = 3200}}}
From above,
{{{c + s = 600}}}
{{{6c + 6s = 3600}}}
And by subtraction:
{{{6c + 6s = 3600}}}
{{{-4c - 6s = -3200}}}
{{{2c = 400}}}
{{{c = 200}}}
and,
{{{c + s = 600}}}
{{{200 + s = 600}}}
{{{s = 400}}}
There were:
200 children
400 students
300 adults
check answer:
(1) {{{4c + 6s + 8a = 5600}}}
(1) {{{4*200 + 6*400 + 8*300 = 5600}}}
(1) {{{800 + 2400 + 2400 = 5600}}}
{{{5600 = 5600}}}
OK