Question 936014
1. let rangos be {{{R}}}, fangos {{{F}}}, and a bango {{{B}}}
{{{R=6F}}}......eq.1

{{{B=15R}}}.....eq.2
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a. Alex has {{{9}}} rangos and {{{4}}} fangos. 
Barbara has {{{7}}} rangos and {{{3}}} fangos. 


{{{Alex=9R+4F}}}..........substitute {{{R}}} from eq.1

{{{Alex=9*6F+4F=54F+4F=58F}}}

{{{Barbara=7R+3F=7*6F+4F=42F+4F=46F}}}

so, {{{Alex+Barbara=58F+46F=104F}}}


2.
easy

A = 41
B = 0
C = 50

proof:

{{{AB+C=50}}}................eq.1
{{{BC+A=41}}}.............eq.2



{{{AB+C=50}}} ......eq.1 ...........solve for{{{ C}}}

{{{C=50-AB}}}..substitute in eg.2


{{{B(50-AB)+A=41}}}......eq.2


{{{50B-AB^2+A=41}}}....solve for {{{A}}}


{{{50B-A(B^2-1)=41}}}


{{{50B=41+A(B^2-1)}}}


{{{50B-41=A(B^2-1)}}}


{{{A=(50B-41)/(B^2-1)}}}


then {{{C=50-((50B-41)/(B^2-1))B}}}


go to

{{{AB+C=50}}}....eq.1 substitute {{{A}}} and {{{C}}}


{{{((50B-41)/(B^2-1))B+50-((50B-41)/(B^2-1))B=50}}}


{{{((50B-41)B+50(B^2-1)-(50B-41)B)/(B^2-1)=50}}}


{{{(50B^2-41B+50B^2-50-50B^2+41B)/(B^2-1)=50}}}


{{{-50/(B^2-1)=50}}}


{{{-50=50(B^2-1)}}}


{{{-1=B^2-1}}}


{{{-1+1=B^2}}}


{{{0=B^2}}} 

=> {{{highlight(B=0)}}}


then {{{C=50-(50B-41)/(B^2-1)B}}}

{{{ C=50-(50*0-41)/(0^2-1)*0}}}

{{{ C=50-(-41/-1)*0}}}

{{{ C=50-0}}}

{{{ highlight(C=50)}}}

and

{{{A=(50B-41)/(B^2-1)}}}

{{{A=(50*0-41)/(0^2-1)}}}

{{{A=-41/-1}}}

{{{highlight(A=41)}}}