Question 904597

let' the number of coins Ken has be {{{x}}} and the number of coins Amy has be {{{y}}} 
Ken and Amy have {{{36}}} coins:

{{{x+y=36}}} ....eq.1

Ken has {{{4}}} more coins than Amy:

{{{x=y+4}}} ...eq.2

go to

{{{x+y=36}}} ....eq.1 and substitute {{{x}}} from eq.2

{{{y+4+y=36}}} ...solve for {{{y}}}

{{{2y=36-4}}}

{{{2y=32}}}

{{{y=32/2}}}

{{{highlight(y=16)}}}......the number of coins Amy has 

now find {{{x}}} 


{{{x=y+4}}} ...eq.2

{{{x=16+4}}} 

{{{highlight(x=20)}}} ....the number of coins Ken has