Question 1140206


let tickets be {{{x}}}that cost​ ${{{10}}} ,{{{y}}}tickets that cost​ ${{{20}}} , and {{{z}}} tickets that cost​ ${{{30}}}

The team has sold{{{ 3203 }}}tickets overall. 

{{{x+y+z=3203}}}....eq.1...solve for {{{z}}}

{{{z=3203-x-y}}}......eq.1a

It has sold {{{216}}} more​ ${{{20}}} tickets than​ ${{{10}}} tickets.

=>{{{y=x+216}}}

 The total sales are ​${{{61 650}}}:


{{{10x+20y+30z=61650}}}....substitute {{{y}}} and {{{z}}}

{{{10x+20(x+216)+30(3203-x-(x+216))=61650}}}.......solve or {{{x}}}

{{{10x+20x+4320+30(3203-x-x-216)=61650}}}

{{{30x+4320+30(3203-2x-216)=61650}}}

{{{30x+4320+96090-60x-6480=61650}}}

{{{93930-30x=61650}}}

{{{93930-61650=30x}}}

{{{32280=30x}}}

{{{x=32280/30}}}

{{{x=1076}}}

go to {{{y=x+216}}}, plug in value for {{{x}}}

{{{y=1076+216}}}

{{{y=1292}}}

go to {{{z=3203-x-y}}}......eq.1a ,plug in value for {{{x}}} and {{{y}}}

{{{z=3203-1076-1292}}}
{{{z=3203-2368}}}

{{{z=835}}}


they sold:

{{{ 1076}}} tickets  that cost​ ${{{10}}}

{{{1292}}} tickets  that cost​ ${{{20}}}

 {{{835}}} tickets  that cost​ ${{{30}}}