Question

divide 47 bwtween A,B,C, so that a may have 10 more than B, AND B 8MORE THEN C.

Answer 1

Let C be x
So B=x+8
And A=x+8+10=x+18
A+B+C=47
x+x+8+x+18=47
3x+26=47
3x=21
x=7=C
15=B
C=25

Answer 2

[0010001111]... Hello User... [1001010101]
Here's your answer...

Let the value of c be x.
Then b has 8 more than c, so b = x+8
a has 10 more than b, so a = x+8+10 = x+18
$x + x + 8 + x + 18 = 47 \\ 3x + 26 = 47 \\ 3x = 47 - 26 = 21 \\ x = \frac{21}{3} = 7$
So c has 7
b has 7+8 = 15
a has 7+18 = 25

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