Simultaneous Linear Equations Problem :-
Sweets are distributed among children. If each child receives 10 sweets, there are 3 left over, but there are 4 sweets less for each to receive 11. Find the number of sweets.
Can you tell me the procedure to solve this question? Thank you :D
Answers
Answered by
3
let there are x children and y sweets
then
10*x + 3 = y
10x - y + 3 = 0 ---------------(1)
11*x = y + 4
11x - y - 4 = 0 --------------------(2)
(2)-(1)
x - 7 = 0
x = 7
y = 10*7 + 3 = 73
total number of sweets = 73
then
10*x + 3 = y
10x - y + 3 = 0 ---------------(1)
11*x = y + 4
11x - y - 4 = 0 --------------------(2)
(2)-(1)
x - 7 = 0
x = 7
y = 10*7 + 3 = 73
total number of sweets = 73
Answered by
1
Let no. of children= n. Then 10n + 3 = 11n -4 ==> n = no. of children = 7.
==> No. of sweets by putting n = 7 in (10n+3) or (11n-4) = 73
==> No. of sweets by putting n = 7 in (10n+3) or (11n-4) = 73
Similar questions