Question 942589

barbecues {{{x}}}
soft drinks {{{y}}}

if a group of students bought {{{10}}} barbecues and {{{12}}} soft drinks at a price of ${{{360}}}, then we have

{{{10x+12y=360}}}.........eq.1

and another group bought {{{5}}} barbecues and {{{6}}} soft-drinks at a price of ${{{120}}}, we have
 

{{{5x+6y=120}}}.........eq.2



solve the system:

{{{10x+12y=360}}}.........eq.1
{{{5x+6y=120}}}.........eq.2
----------------------------------subtract 2 from 1

{{{10x+12y-5x+6y=360-120}}} 

{{{5x+6y=240}}} 

{{{5x=240-6y}}} 

{{{x=(240-6y)/5}}} ....substitute in 2

{{{5(240-6y)/5+6y=120}}}.........eq.2

{{{240-6y+6y=120}}}...since {{{-6y+6y=0}}}

{{{240<>120}}} => lines are parallel and there is no solution; this system is inconsistent

{{{10x+12y=360}}}.........eq.1
{{{5x+6y=120}}}.

{{{ graph( 600, 600, -10, 10, -10, 50, -10x/12+36, -5x/6+20) }}}

so, we cannot determine what is the price of a single barbecue and soft-drink