Solving a maths question

Ming, Nordin and Owen had 70 toy cars altogether. After Nordin gave Owen 6 toy cars and Ming lost 10 toy cars, the 3 boys ended up with the same number of you cars. Find the number of toy cars that Owen had at first

Well if ming lost 10 that leaves you with 60 toys split 3 ways which is 20 each but since nor din gave away 6 to Owen your answer will be 14 Owen started with 14 toys and nor din had 26 while ming had 30

Let the number of toys be m, n and o respectively.

so, after the losses and exchanges we have:

Ming --> m - 10

Nordin --> n - 6

Owen --> o + 6

so, m - 10 = n - 6....n - 6 = o + 6.....and m - 10 = o + 6

i.e. m - n = 4...(1)

=> n - o = 12...(2)

=> m - o = 16..(3)

Now, m + n + o = 70

so, m + (m - 4) + (m - 16) = 70

=> 3m - 20 = 70

=> 3m = 90

i.e. m = 30 => n = 26 and o = 14

so, at the beginning, Ming has 30 cars, Nordin has 26 cars and Owen has 14 cars

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Total number of cars = 70
M + N + O = 70

Turn the second sentence into an equation also
N - 6 = O + 6 = M - 10

Rearrange:
N = O + 12
M = O + 16

Substitute into first equation:
O + 16 + O + 12 + O = 70

Solve for O:
3O + 28 = 70
3O = 42
O = 14

Owen initially had 14 cars

Ming = x cars

Nordin = y cars

Owen = 70 - x - y

Ming lost 10 remaining, x - 10
Nordin gave Owen 6 remaining y - 6
Owen now has 76 - x - y

Now x - 10 = y - 6 = 76 - x - y

x - 10 = y - 6
x - y = 10 -6
x - y = 4, y = x - 4

x - 10 = 76 - x - y
x - 10 = 76 - x - (x - 4)
x - 10 = 76 - x - x + 4
x - 10 = 80 - 2x
3x = 90
x = 30

y = 30 - 4 = 26

Owen first had, 70 - 30 - 26 = 14