Question 907836

let's numbers be {{{x}}} and {{{y}}}


If the difference between two numbers is {{{6}}},we have

{{{x-y=6}}}......eq.1

 and if the difference of their squares is {{{180}}}, 

{{{x^2-y^2=180}}}.............eq.2

you got system to solve:


{{{x-y=6}}}......eq.1
{{{x^2-y^2=180}}}.............eq.2
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{{{x-y=6}}}......eq.1...solve for {{{x}}}

{{{x=y+6}}}......substitute in eq.2


{{{(y+6)^2-y^2=180}}} .............eq.2...solve for {{{y}}}

{{{y^2+12y+36-y^2=180}}}

{{{cross(y^2)+12y+36-cross(y^2)=180}}}

{{{12y=180-36}}}

{{{12y=144}}}

{{{y=144/12}}}

{{{y=12}}}

now find {{{x=y+6}}}

{{{x=12+6}}}

{{{x=18}}}

check:

{{{x-y=6}}}=>{{{18-12=6}}}=>{{{6=6}}}
{{{x^2-y^2=180}}}=> {{{18^2-12^2=180}}}=>{{{324-144=180}}}=>{{{180=180}}}