Question

the sum of two number is 184. One third of first number exceeds one seventh of another number by 8. Find the numbers.

Answer 1

hi
a+b=184
a/3-b/7=8
7a-3b=8*21,,,,,,,,,(1)
7a-3b=168and a+b=184 now multiply this by 3
3a+3b=184*3=552,,,,,,,,,,(2)
now add one and two equation
10a =168+552=720
so a =720/10=72
and b=184-72=112

so answer is a=72
b =112


hope it helps,,,,,,,,

Answer 2

let the 1st number be x and 2nd number be y
x+y =184- (i)
a/q
x/3-y/7=8
(7x-3y)/21=8
7x-3y=168 -(ii)
so,
x+y=184 -i
7x-3y=168 -ii
on multiplying eq i by 3 and ii by 1 we get
3x+3y= 552
7x-3y= 168
on adding these 2 eq we get
10x = 720
x= 72
y=112 .