A B C X Y Z X + 350 Y − 350 Z X + 350 Y + 350 Z − 700 X + 140 Y + 350 Z − 490 \def\arraystretch{1.5}
\begin{array}{c:c:c}
A & B & C \\ \hline
X & Y & Z \\
\hdashline
X+350 & Y-350 & Z
\\ \hdashline
X+350&Y+350&Z-700
\\ \hdashline
X+140& Y+350&Z-490
\end{array} A X X + 350 X + 350 X + 140 B Y Y − 350 Y + 350 Y + 350 C Z Z Z − 700 Z − 490
B B B loses Php 350 of his money to A A A : A A A has X + 350 X+350 X + 350 and B B B has Y − 350. Y-350. Y − 350.
A A A now has twice as much as what is left with B B B : X + 350 = 2 ( Y − 350 ) . X+350=2(Y-350). X + 350 = 2 ( Y − 350 ) .
C C C loses Php 700 to B B B : C C C has Z − 700 Z-700 Z − 700 and B B B has Y + 350 Y+350 Y + 350 .
C C C now has only one-third as much money as B B B would then have: Z − 700 = 1 3 ( Y + 350 ) . Z-700=\frac{1}{3}(Y+350). Z − 700 = 3 1 ( Y + 350 ) .
If A A A has loses Php 210 to C C C , C C C will have as much as money as A A A would have left: A A A has X + 140 X+140 X + 140 and C C C has Z − 490 Z-490 Z − 490 and X + 140 = Z − 490. X+140=Z-490. X + 140 = Z − 490.
{ X + 350 = 2 ( Y − 350 ) Z − 700 = 1 3 ( Y + 350 ) X + 140 = Z − 490 \begin{cases}
X+350=2(Y-350)
\\
Z-700=\frac{1}{3}(Y+350)
\\
X+140=Z-490
\end{cases} ⎩ ⎨ ⎧ X + 350 = 2 ( Y − 350 ) Z − 700 = 3 1 ( Y + 350 ) X + 140 = Z − 490 { X = 2 Y − 1050 Y = 3 Z − 2450 Z = X + 630 \begin{cases}
X=2Y-1050
\\
Y=3Z-2450
\\
Z=X+630
\end{cases} ⎩ ⎨ ⎧ X = 2 Y − 1050 Y = 3 Z − 2450 Z = X + 630 { X = 6 X − 2170 Y = 3 X − 560 Z = X + 630 \begin{cases}
X=6X-2170
\\
Y=3X-560
\\
Z=X+630
\end{cases} ⎩ ⎨ ⎧ X = 6 X − 2170 Y = 3 X − 560 Z = X + 630 { X = 434 Y = 742 Z = 1064 \begin{cases}
X=434
\\ Y=742\\ Z=1064\end{cases} ⎩ ⎨ ⎧ X = 434 Y = 742 Z = 1064
Answer: A A A - 434 , 434, 434 , B B B - 742 , 742, 742 , C C C - 1064. 1064. 1064.
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