The larger number #L# is #11# more or #+11# than twice the smaller #S# number or #2 xx S#.
That means #L = 2S + 11#.
Now #3# times the larger or #3 xx L# is #9# more or #+ 9# than #7# times the smaller or #7S#.
That means #3L = 7S + 9#.
We can multiply both sides of the first equation by #3# so it will match the second equation for the #L# term.
#3L = 6S +33#
We can now substitute the value of #3L# into the second equation:
#6S + 33 = 7S + 9#
Subtract #S# values from the right side and number values from the right side to bring them across the #=# sign.
#-S = -24#
#S = 24#
Using the first equation to solve for #L#;
#L = 2S + 11#
#L = 2(24) + 11#
#L = 59#
To check, substitute the values back into the second equation:
#3L = 6S + 9#
#177 = 168 + 9#
#177 = 177#
