Question 1077493

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

if the sum of two numbers is {{{33}}}, we have

{{{x+y=33}}} .............solve for {{{x}}} 

{{{x=33-y}}} ....eq.1
 
and, if their product is {{{270}}}, we have

{{{x*y=270}}} ...........solve for {{{x}}} 

{{{x=270/y}}}.......eq.2

since left sides of eq.1 and eq.2 are same, right sides are equal too

{{{33-y=270/y}}}.......solve for {{{y}}}

{{{(33-y)y=270}}}

{{{33y-y^2=270}}}

{{{0=y^2-33y+270}}}...factor

{{{0 = y^2-15y -18y +270}}}

{{{0 = (y^2 -18y) -(15y-270)}}}

{{{0 = y(y -18) -15(y-18)}}}

{{{0 = (y -18)(y -15)}}}

solutions:

if {{{0 = (y -18)}}}=>{{{y=18}}}
if {{{0 = (y -15)}}}=>{{{y=15}}}

now find {{{x}}}

{{{x=33-y}}} ....eq.1 if {{{y=18}}}
{{{x=33-18}}}
{{{x=15}}}

{{{x=33-y}}} ....eq.1 if {{{y=15}}}
{{{x=33-15}}}
{{{x=18}}}

so, your numbers are {{{15}}} and {{{18}}}