Guide 4 Foundations of Math Ch. 8 Drill 3. Drill 3. #2.

[anon]

Drill 3: Solve the following inequalities.

2. x/3 +8 < x/2

In the answer it multiplies both sides by 6 to cancel out the bottom fractions. Can you please explain? When I solve the problem I multiply the 3 on both sides to cancel out the three and then multiply by 2 on both sides to cancel out the fraction on the right side. I see why the 6 was used since they both multiply by 6. What I don't see if how I can recognize when to multiply by a number that's not already in the problem while knowing it's still an approved number and not something that's changing the equation. Thanks!

[tim]

You can totally multiply by 3 and by 2. That does the same thing as multiplying by 6. You save yourself a few seconds if you recognize this and just start with the 6, but your method is totally fine as well..