Lets say two positive integers are #x# and #y#.
So #x+y = 10# and #x xx y = 21#
We have two equations here:
#x + y = 10# ----> equation 1
#xy = 21# ----> equation 2
From equation 1, make #y# the subject.
#x + y = 10#
#y = 10-x# --- substitute #y# in equation 2
#xy = 21# ---- Equation 2
#x(10-x) = 21
#10x-x^2 = 21#
Re-arranging the equation, we get:
#-x^2+10x-21=0#
Lets solve for #x# now:
#-x^2+10x-21=0#
Multiply by (#-1#) throughout and we get:
#x^2-10x+21=0#
Factors are #-3# and #-7# as #-3 + -7=10# and #-3 xx (-7) = 21#
#x^2-10x+21=0#
#x^2-3x-7x+21=0#
#x(x-3)-7(x-3)=0#
#(x-7)(x-3)=0#
Hence, #x=7# or #x=3#
Substitute value of #x#in equation 1
Take #x=7#
#7 + y = 10#
#y=10-7#
#y=3# -----> First value of #y#
Take #x=3#
#3 + y = 10#
#y=10-3#
#y=7# ------> Second value of #y#
Therefore Answer is:
#x=3# or #x=7#
#y=7# or #y=3#
And all of them are positive.
