Ravi tests whether $y=6$ is a solution of $4y-5=23$. He works out the LHS and gets 19. What should he conclude?

A value is a solution only when it makes the LHS exactly equal to the RHS. Here $4(6)-5=19$ while the RHS is 23, so $y=6$ fails. Solving properly: $4y=28$ → $y=7$, and $4(7)-5=23$ ✓. The trap is treating 'I managed to work out the LHS' as 'the value works'; the RHS is the constant 23 and has no $y$ to substitute into.