Answer :
let x and y be the integers
we are told that 4x = 3y + 23 EQN (1)
and x = y + 1 EQN (2)
We have two unknowns and two equations meaning we can solve this system.
Substituting (2) into (1) we get
4(y+1) = 3y + 23
which solves to y = 19 which in turn means x = 20
we are told that 4x = 3y + 23 EQN (1)
and x = y + 1 EQN (2)
We have two unknowns and two equations meaning we can solve this system.
Substituting (2) into (1) we get
4(y+1) = 3y + 23
which solves to y = 19 which in turn means x = 20