[tex]The\ given\ situation\ is\ a\ \bold{Trial\ and\ Error}\ equation. \\ \\ \bold{Equation\ to\ be\ solved:} \\ \\ Let\ \bold{x,\ y\ and\ z}\ be\ the\ unknown\ numbers. \\ \\ x+y+z=30 \\ \\ \bold{Solution:} \\ \\ x+y+z=30 \\ \\ Let\ the\ unknown\ numbers\ \bold{ \sum \to \ 30} \\ \\ If\ odd+odd=even \\ If\ odd+odd+even=even \\ If\ odd+odd+odd=odd[/tex]
[tex]Since,\ 30\ is\ an\ even\ number,\ then\ we\ must\ follow\ the\ rules \\ odd+odd+even=even. \\ \\ There,\ is\ only\ one\ even\ number\ and\ that\ is\ 4, \\ therefore,\ variable\ \bold{z}\ solved. \\ \\ \bold{Equation:} \\ x+y+4=30 \\ \\ x+y=30-4 \\ \\ \bold{x+y=26} \\ \\ Therefore,\ \bold{x+y\ must\ equal\ to\ 26.}[/tex]
[tex]Think\ of\ a\ number\ from\ the\ given\ that\ \bold{equals\ to\ 26.} \\ \\ \bold{\{15\ and\ 11\}=26} \\ \\ Therefore,\ the\ value\ of\ x\ and\ y\ are\ 15\ and\ 11. \\ Hence,\ the\ value\ of\ the\ equation\ is; \\ \\ x+y+z=30\implies \bold{15+11+4=30} \\ \\ \bold{Check;} \\ x+y+z=30 \\ \\ 15+11+4=30 \\ \\ \bold{30=30}\ \underline{TRUE}[/tex]
[tex]Therefore,\ the\ value\ of\ the\ three\ unknown\ numbers\ are \\\bold{11,\ 15\ and\ 4}\ in\ any\ orders.[/tex]
