Question 701258

let the cost of  pennants be {{{x}}} and the cost of balls {{{y}}}:

if the cost of six pennants and nine balls is ${{{48}}} we can write it as

{{{6x+9y=48}}}......eq. 1


if the cost of three balls is ${{{1}}} more than the cost of one pennant, than we write it as

{{{3y=x+1}}}....eq. 2

now we have system to solve:

{{{6x+9y=48}}}......eq. 1
{{{3y=x+1}}}....eq. 2
__________________________

start with {{{3y=x+1}}}....eq. 2 and solve for {{{y}}}

{{{y=(x+1)/3}}}.....substitute it in eq. 1

{{{6x+9(x+1)/3=48}}}......solve for {{{x}}}

{{{6x+cross(9)3(x+1)/cross(3)=48}}}

{{{6x+3(x+1)=48}}}

{{{6x+3x+3=48}}}

{{{9x=48-3}}}

{{{9x=45}}}

{{{x=45/9}}}

{{{highlight(x=5)}}}....the cost of  pennants

now find {{{y}}}

{{{y=(5+1)/3}}}

{{{y=6/3}}}

{{{highlight(y=2)}}}......the cost of balls