We simplify this inequality similarly to how we would simplify an equation

\(\displaystyle \dpi{100} \small 3x+8-8< 35-8\)

\(\displaystyle \dpi{100} \small \frac{3x}{3}< \frac{27}{3}\)

Thus \(\displaystyle \dpi{100} \small x< 9\)

In order to find the solution set, we solve \(\displaystyle 2x+12>42\) as we would an equation:

\(\displaystyle 2x+12>42\)

\(\displaystyle 2x>30\)

\(\displaystyle x>15\)

Therefore, the solution set is any value of \(\displaystyle x>15\).

The inequality that is presented in the problem is:

\(\displaystyle 8-7x\leq11x+4\)

Start by moving your variables to one side of the inequality and all other numbers to the other side:

\(\displaystyle -7x-11x\leq4-8\)

\(\displaystyle -18x\leq-4\)

Divide both sides of the equation by \(\displaystyle -18\). Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

\(\displaystyle x\geq\frac{-4}{-18}\)

Reduce:

\(\displaystyle x\geq\frac{2}{9}\)

The only answer choice with a value greater than \(\displaystyle \frac{2}{9}\) is \(\displaystyle \frac{4}{9}\).

To get the lowest value, you need the lowest numerator and the highest denominator.  That would be \(\displaystyle \frac{2}{10}\) or reduced to be \(\displaystyle \frac{1}{5}\).  For the highest value, you need the highest numerator and the lowest denominator.  That would be \(\displaystyle \frac{12}{9}\) or \(\displaystyle \frac{4}{3}\).

The inequality \(\displaystyle |- 5 x + 32| \le 38\) can be rewritten as the three-part inequality

\(\displaystyle -38 \le - 5 x + 32 \le 38\)

Isolate the \(\displaystyle x\) in the middle expression by performing the same operations in all three expressions. Subtract 32 from each expression:

\(\displaystyle -38 - 32 \le - 5 x + 32 - 32 \le 38 - 32\)

\(\displaystyle -70 \le - 5 x \le 6\)

Divide each expression by \(\displaystyle -5\), switching the direction of the inequality symbols:

\(\displaystyle -70\div (-5) \ge - 5 x\div (-5) \ge 6 \div (-5)\)

\(\displaystyle 14 \ge x \ge -1.2\)

This can be rewritten in interval notation as \(\displaystyle [-1.2, 14 ]\).

In an absolute value inequality, the absolute value expression must be isolated first, as follows:

\(\displaystyle - 12 - | -4y + 52 | > 32\)

Adding 12 to both sides:

\(\displaystyle - 12 - | -4y + 52 | + 12 >32+ 12\)

\(\displaystyle - | -4y + 52 | >44\)

Multiplying both sides by \(\displaystyle -1\), and switching the inequality symbol due to multiplication by a negative number:

\(\displaystyle - | -4y + 52 | \cdot (-1 ) < 44 \cdot (-1 )\)

\(\displaystyle | -4y + 52 | < -44\)

We do not need to go further. An absolute value expression must always be greater than or equal to 0; it is impossible for the expression \(\displaystyle | -4y + 52 |\) to be less than any negative number. The inequality has no solution.

In general, when we take the absolute value of a quantity, we can represent it as either itself, or as its additive inverse.

In other words, |x| = x (if x > 0) or |x| = –x (if x < 0).

Therefore, we can represent |4 – 2x| as either 4 – 2x or as –(4 – 2x). We must consider both of these cases and solve for x. Let’s us first consider the case that |4 – 2x| = 4 – 2x.

4 – 2x < 5

We can add 2x to both sides

4 < 2x + 5

Then subtract 5 from both sides.

–1 < 2x

Divide by 2.

–1/2 < x

x > –0.5

Now, we consider the case that |4 – 2x | = –(4 – 2x).

–(4 – 2x) < 5

Multiply both sides by negative one. Remember, whenever we multiply or divide an inequality by a negative number, we must flip the inequality sign.

4 – 2x > –5

Add 2x to both sides.

4 > 2x – 5

Add 5 to both sides.

9 > 2x

Divide by 2,

9/2 > x

4.5 > x

This means that the values of x that satisfy the original quality must be greater than -0.5 AND less than 4.5.

The question asks us to find the sum of the integer values of x that satisfy the inequality. The only integers between -0.5 and 4.5 are the following: 0, 1, 2, 3, and 4.

The sum of 0, 1, 2, 3, and 4 is 10.

The answer is 10.

There are 3 solutions: 0, 7, 7.

The correct answer is 2 distinct solutions, however, because 7 occurs twice.  So the two distinct solutions are 0 and 7.

No common terms cancel out, and this isn't a difference of squares. 

Let's move the 48 to the other side: x² = 48

Now take the square root of both sides: x = √48 or x = –√48. Don't forget the second (negative) solution! 

Now √48 = √(3*16) = √(3*4²) = 4√3, so the answer is x = 4√3 or x = –4√3.

If \(\displaystyle \dpi{100} \small |x+3| = 10\), then either

\(\displaystyle \dpi{100} \small x+3 = 10\) or \(\displaystyle \dpi{100} \small x+3 = -10\).  

These two equations yield \(\displaystyle \dpi{100} \small 7\) and \(\displaystyle \dpi{100} \small -13\) as answers.  

\(\displaystyle \dpi{100} \small 7+(-13)=-6\)