Question 483085

let one number be {{{x}}} and another {{{y}}}

given:

{{{x=(2/3)y}}}....1

{{{x+y=60}}}......2
------------------------solve this system


{{{x+y=60}}}......2...replace {{{x}}} with {{{(2/3)y}}} from equation 1


{{{(2/3)y+y=60}}}....both sides multiply by {{{3}}}


{{{3(2/3)y+3y=60*3}}}


{{{cross(3)(2/cross(3))y+3y=180}}}


{{{2y+3y=180}}}


{{{5y=180}}}


{{{y=180/5}}}


{{{highlight(y=36)}}}

now find {{{x}}}


{{{x=(2/3)y}}}....1...plug in {{{y}}} value


{{{x=(2/3)36}}}


{{{x=(2/cross(3))cross(36)12}}}


{{{x=2*12}}}


{{{highlight(x=24)}}}


check:

{{{x+y=60}}}......2.


{{{24+36=60}}}


{{{60=60}}}