# How do you solve 3x-1<2  or -5x<-20 ?

May 19, 2015

With the first inequality, first add $1$ to both sides to get:

$3 x < 3$

Then divide both sides by $3$ to get:

$x < 1$

With the second inequality, first add $5 x$ to both sides to get:

$0 < 5 x - 20$

Then add $20$ to both sides to get:

$20 < 5 x$

Finally divide both sides by $5$ to get:

$4 < x$

Putting these together we have:

$x < 1$ or $x > 4$

Notice what I did not do. I did not divide both sides of $- 5 x < - 20$ by $- 5$. If you divide both sides by a negative number then you must also reverse the inequality, which would have given us $x > 4$ in shorter order, but might have required more explanation.

In general you can perform any of the following operations and preserve the truth of an inequality:

(1) Add or subtract the same value on both sides.
(2) Multiply or divide both sides by the same positive value.
(3) Multiply or divide both sides by the same negative value and reverse the inequality.