Let the numbers be x and y
x be the larger; y be the smaller
x + y = 23
Double the larger the number exceeds three times the smaller by sixteen
2x = 3y + 16
Let's rearrange the first equation to find the value of x
x + y = 23
Transpose y
x = 23 - y
Now substitute the value of x to the 2nd equation
2x = 3y + 16
2(23 - y) = 3y + 16
Multiply
46 - 2y = 3y + 16
Transpose 3y and 46
-2y - 3y = 16 - 46
Add like sign and subtract unlike sign
-5y = -30
Divide by -5 both sides
-5y/-5 = -30/-5
Therefore
y = 6
Let us now find for x
x = 23 - y
Substitute value of y
x = 23 - 6
Therefore
x = 17
Let us check by substituting the value of x and y to the second equation
2x = 3y + 16
2(17) = 3(6) + 16
Multiply
34 = 18 + 16
Add
34 = 34
I hope it helps
