|4x-7|<21
I seem to be getting very confused with absolute values. I know that if anyhting is negative inside the absolute value, it becomes positive but can someone explain the steps to answer above. Thanks
you solve it as you normally would if you had an equal sign, except that, when using absolute values, whenever you are sending something negative multiplying or dividing to the other side of the sign, the sign changes, so if it was < it becomes > and vice versa.
In this case you are not sending anything negative that multiplies or divides, so the end result is:
4x<21+7
4x<28
x<28/4
x<7
so as you can see, x has to be lower than 7 for the statement to be true.
EDIT:
I forgot what Gusbob posted, there is a range for x, check his post below, he has the right way to do it.
This can also be written as acompound inequality
if $\displaystyle |x|< 3$ then $\displaystyle -3< x < 3$ is the solution
if we apply this to the above problem we get
$\displaystyle |4x-7|<21 \iff -21 < 4x-7 <21$ add 7 to all sides
$\displaystyle -21+7<4x-7 +7 < 21+7 \iff -14 <4x < 28$ divide by 4
$\displaystyle \frac{-14}{4} < x < 7 \iff \frac{-7}{2} < x < 7$
so our solution is $\displaystyle (-\frac{7}{2},7)$