How do you solve $12<8+x$ ?

Question

4069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-21T20:29:30

$x>4$

Subtract $8$ from both sides to get

$12-8 < cancel(8)+x-cancel(8)$

The left hand side becomes $12-8=4$ and so we have

$4<x$

which is the same as

$x>4$

How do you solve 12<8+x12<8+x?