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the sum of two odd numbers is 28.the larger number is sixteen more than one-third the smaller number.find the numbers

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Elensapid
[tex]a+b=28 \\ \\ a=16+ \frac{b}{3} \\ \\ ^{3)}16+ \frac{b}{3}+^{3)}b=^{3)}28 \\ \\ \frac{48}{3}+ \frac{b}{3}+ \frac{3b}{3}= \frac{84}{3}|*3 \\ \\ 4b+48=84 \\ \\ 4b=36 \\ \\ \boxed{b=9} \\ \\ a=28-9 \\ \\ \boxed{a=19} [/tex]
Jeremiaid
x + y = 28
16 + [tex] \frac{x}{3} [/tex] = y
Multiply 3 to all sides

48 + x = 3y
Transpose

48 + x - 3y = 0
Transpose more

x - 3y = -48

Subtract the 2nd equation from the first equation, like this:
  
     x + y =  28
-    x - 3y = -48
     0 + 4y = 76
    
     4y = 76
Divide 4 to both sides

y = 19

Substitute value of y in the first equation

x + y = 28
x + 19 = 28

Transpose

x = 28 - 19

x= 9
y= 19

Therefore, the numbers are 9 and 19.