solve each inequality.?

1) 5x + 2 <= 17

Isolate x. Subtract 2 from both sides of the inequality, to get

5x <= 15

Divide 5 both sides,

x <= 3

In interval notation, this is written as

(-infinity, 3]

2) For the second inequality, it would be the WRONG step to multiply both sides by (5x + 6), so PLEASE nobody under me do it.

Let me show you instead how it is solved.

1/(5x + 6) > -3

First, move the -3 to the left hand side.

1/(5x + 6) + 3 > 0

Next, merge as a single fraction.

1/(5x + 6) + 3(5x + 6)/(5x + 6) > 0

(1 + 3(5x + 6)) / (5x + 6) > 0

Simplify the numerator.

(1 + 15x + 18) / (5x + 6) > 0
(15x + 19) / (5x + 6) > 0

Obtain our critical values by noting what makes the left hand side equal to 0 or what makes the left hand side undefined.

It is 0 when the numerator is 0, i.e. when
15x + 19 = 0, or x = -19/15
It is undefined when the denominator is undefined, i.e. when
5x + 6 = 0, or x = -6/5 = -18/15

Make a number line consisting of these critical values.

. . . . . . . . . . (-19/15) . . . . . . . . . .(-18/15) . . . . . . . . . . . . . . .

This number line is your sign diagram. You want to test a single value in each region, plug into (15x + 19) / (5x + 6), and determine if the result is positive or negative.

For x < -19/15, test x = -100. Then

(15x + 19) / (5x + 6) = (negative) / (negative) = positive.
Mark the region as positive.

. . . .{+} . . . . (-19/15) . . . . . . . . . .(-18/15) . . . . . . . . . . . . . . .

For -19/15 < x < -18/15, it's a tricky area to test... if we make them both over 30, we get
-38/30 < x < -36/30

So let's test x = -37/30.

(15x + 19) / (5x + 6) = ( 15(-37/30) + 19) / (5(-37/30) + 6)
= (-37/2 + 19) / (-37/6 + 6)
= (-18.5 + 19) / (-37/6 + 36/6)
= (positive) / (negative)
= negative.
Mark the region as negative.

. . . .{+} . . . . (-19/15) . . . .{-} . . . .(-18/15) . . . . . . . . . . . . . . .

For x > -18/15, test x = 1000000000. Then
(15x + 19) / (5x + 6) = (positive) / (positive)
= positive.
Mark the region as positive.

. . . .{+} . . . . (-19/15) . . . .{-} . . . .(-18/15) . . . . .{+}. . . . . . . .

Look back at what the question is asking:

(15x + 19) / (5x + 6) > 0

So it is asking when the inequality is greater than 0, i.e. positive. Look at the regions themselves, it is positive when

(x < -19/15) OR (x > -18/15)

In interval notation, this is written as

(-infinity, -19/15) U (-18/15, infinity)

And that's your answer.

1)
5x + 2 ≤ 17
5x ≤ 17 - 2
x ≤ 15/5
x ≤ 3

2)
1/(5x + 6) > -3
1 > -3(5x + 6)
1 > -15x - 18
-15x < 1 + 18
x > 19/-15
x > -19/15

Question 1
5x + 2 ≤ 17
5x ≤ 15
x ≤ 3

Question 2
As given question MUST be read as :-
(1/5) x + 6 > - 3
x + 30 > - 15
x > - 45