Answer to Question #332688 in Linear Algebra for Monching

Question

A, B and C are on a betting game. B loses Php 350 of his money to A. As a result, A now has twice as much as what is left with B. Then, C loses Php 700 to B. As a consequence, C now has only one-third as much money as B would then have. If A has loses Php 210 to C, C will have as much as money as A would have left. How much did each have at the start?

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c} A & B & C \\ \hline X & Y & Z \\ \hdashline X+350 & Y-350 & Z \\ \hdashline X+350&Y+350&Z-700 \\ \hdashline X+140& Y+350&Z-490 \end{array}$

$B$ loses Php 350 of his money to $A$ : $A$ has $X+350$ and $B$ has $Y-350.$

$A$ now has twice as much as what is left with $B$: $X+350=2(Y-350).$

$C$ loses Php 700 to $B$ : $C$ has $Z-700$ and $B$ has $Y+350$ .

$C$ now has only one-third as much money as $B$ would then have: $Z-700=\frac{1}{3}(Y+350).$

If $A$ has loses Php 210 to $C$ , $C$ will have as much as money as $A$ would have left: $A$ has $X+140$ and $C$ has $Z-490$ and $X+140=Z-490.$

$\begin{cases} X+350=2(Y-350) \\ Z-700=\frac{1}{3}(Y+350) \\ X+140=Z-490 \end{cases}$ $\begin{cases} X=2Y-1050 \\ Y=3Z-2450 \\ Z=X+630 \end{cases}$ $\begin{cases} X=6X-2170 \\ Y=3X-560 \\ Z=X+630 \end{cases}$ $\begin{cases} X=434 \\ Y=742\\ Z=1064\end{cases}$

Answer: $A$ - $434,$ $B$ - $742,$ $C$ - $1064.$