Question 1176314


{{{7ax + 2y = 20 }}}....eq.1
{{{ax + 3y = -8}}}......eq.2
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start with 

{{{7ax + 2y = 20 }}}....solve for {{{y}}}

{{{2y = 20 -7ax}}} 

{{{y = 10 -7ax/2 }}}.......a)

then do same with

{{{ax + 3y = -8}}}

{{{ 3y = -8-ax  }}}

{{{y = -8/3-ax/3}}}.............b)

from a) and b) we have

{{{10 -7ax/2 =-8/3-ax/3}}} ........solve for {{{x}}}

{{{10 +8/3 =7ax/2-ax /3}}}...........both sides multiply by {{{6}}}

{{{60 +16 =21ax-2ax }}}

{{{76 =19ax }}}

{{{76/19a=x}}} 

{{{x=4/a}}}


go to

{{{y = -8/3-ax/3}}}......substitute {{{x}}}

{{{y = -8/3-a(4/a)/3}}}

{{{y = -8/3-4/3}}}

{{{y = -12/3}}}

{{{y = -4}}}

solution:({{{4/a}}}, {{{ -4}}})