Question 1123684
let's hockey tickets be {{{x}}} and basketball tickets {{{y}}}


if given that two hockey tickets and one basketball ticket cost ${{{110.40}}}, we have

{{{2x+y=110.40}}}....eq.1

 if one one hockey ticket and two basketball tickets cost ${{{106.32}}}, we have

{{{x+2y=106.32}}}....eq.2

now solve this system

start with {{{2x+y=110.40}}}....eq.1, solve for {{{y}}}

{{{y=110.40-2x}}}....substitute in {{{x+2y=106.32}}}....eq.2

{{{x+2(110.40-2x)=106.32}}}....solve for {{{x}}}

{{{x+220.80-4x=106.32}}}

{{{220.80-3x=106.32}}}

{{{220.80-106.32=3x}}}

{{{114.48=3x}}}

{{{x=114.48/3}}}

{{{x=38.16}}} -> one hockey tickets costs {{{highlight(38.16)}}}


now, go to {{{y=110.40-2x}}}, plug in {{{38.16}}} for {{{x}}}

{{{y=110.40-2*38.16}}}

{{{y=110.40-76.32}}}

{{{y=34.08}}}-> one basketball tickets costs {{{highlight(34.08)}}}