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Let $x$ be the number Eric thought of. He adds $1,$ which gives $x+1.$ This is the sum which he divides by $2,$ leading to $2x+1 .$
This is multiplied by $3$ which gives $2x+1 ⋅3$. Finally, he subtracts $4:$
$2x+1 ⋅3−4.$
This expression should be equal to $5,$ which gives us an equation that we can solve for $x.$

$2x+1 ⋅3−4=5$

AddEqn$LHS+4=RHS+4$

$2x+1 ⋅3=9$

DivEqn$LHS/3=RHS/3$

$2x+1 =3$

MultEqn$LHS⋅2=RHS⋅2$

$x+1=6$

SubEqn$LHS−1=RHS−1$

$x=5$

The number Eric thought of was $5.$