how do you do this?

[ejames2]

|x-7|<0.3


How do you solve this inequality using an algebraic method? how do you make an labeled number line for this?[/u][/code]

 

[skeeter]

|x - 7| < 0.3

if (x - 7) is positive, then x - 7 < 0.3

if (x - 7) is negative, then -(x - 7) < 0.3, or x - 7 > -0.3

the two conditionals above may be written as an interval ...

-0.3 < x - 7 < 0.3

add 7 to every term ...

6.7 < x < 7.3

 

[soroban]

Hello, ejames2!

\(\displaystyle \;|x\,-\,7|\;<\;0.3\)

If we have: \(\displaystyle \,|f(x)|\:<\:c\), then \(\displaystyle f(x)\) is between \(\displaystyle \,-c\,\) and \(\displaystyle \,+c\).

We can write: \(\displaystyle \:-0.3 \:< \:x\,-\,7 \:< \:0.3\)

Add 7 to all three sides: \(\displaystyle \:\fbox{6.7 \:<\:x\:<\:7.3}\)


So \(\displaystyle x\) is between 6.7 and 7.3 . . . can you "see" the interval?

\(\displaystyle \;\;\; - - + - - - - - - \circ = = = = = = = = \circ - -\)
. . . . . .\(\displaystyle 0\) . . . . . . \(\displaystyle 6.7\). . . . . . . . . . .\(\displaystyle 7.3\)

There are "open circles" at 2.7 and 3.3
. . because those values are not included.



If the endpoint were included, say, \(\displaystyle 1\:\leq\:x\:\leq 3\), we'd have:

\(\displaystyle \;\;\;- - + - - - - - \bullet = = = = = = = \bullet - -\)
. . . . . \(\displaystyle 0\) . . . . . .\(\displaystyle 1\) . . . . . . . . . .\(\displaystyle 3\)