Respuesta :

We have the next inequality:

[tex]l3y+7l=10[/tex]

Apply the absolute value properties:

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

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.

Ver imagen IdanG266348
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