Question

How do I solve interval notation and graph it? l3y+7I < 10$|3y + 7| \ \textless \ 10$

Answer 1

We have the next inequality:

$l3y+7l=10$

Apply the absolute value properties:

$\text{lal}=0\text{ so }-aWhich means :[tex]-10<3y+7<10$

3y+7>-10 and 3y+7<10

Solve the inequalities:

for 3y+7>-10

3y>-10-7

y>-17/3

Now, for 3y+7<10

3y<10-7

3y< 3

y< 3/3

y< 1

So the solution is -17/3 < y < 1

Interval notation (-17/3, 1)

Now, the graph:

x can take any real value

y has the interval (-17/3, 1)

-17/3 = -5.6

on y -axis -5.6 and 1 are asymptotes.