For this problem, we must take into account the absolute value.

First, we solve for 2x – 2 > 20.  But we must also solve for 2x – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).  

First step:

2x – 2 > 20

2x > 22

x > 11

Second step:

2x – 2 < –20

2x < –18

x < –9

Therefore, x > 11 and x < –9.

A possible value for x would be –10 since that is less than –9.  

Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.

Move +5 using subtraction rule which will give you. 

Divide both sides by 2 (using division rule) and you will get  which is the same as

Subtract 5 from both sides of the inequality:

Multiply both sides by 5:

Therefore only I must be true.

Solve for both x – 3 < 2 and –(x – 3) < 2.

x – 3 < 2 and –x + 3 < 2

x < 2 + 3 and –x < 2 – 3

x < 5 and –x < –1

x < 5 and x > 1

The results are x < 5 and x > 1.

Combine the two inequalities to get 1 < x < 5

Manipulate the inequality until  is on a side by itself:

For this equation,  must be less than 6. Find the answer choice with values all less than 6. In this case, it will be -1, 4, and 5.

Subtract  from both sides:

Subtract 7 from both sides:

Divide both sides by :

Remember to switch the inequality when dividing by a negative number:

Since is not an answer, we must find an answer that, at the very least, does not contradict the fact that  is less than (approximately) 4.67.  Since any number that is less than 4.67 is also less than any number that is bigger than 4.67, we can be sure that  is less than 5.

The median weight of a box of cereal is 360 grams. This should be an allowable value of w. Substituting 360 for w into each answer choice, the only true results are:

and:

Notice that any positive value for w satisfies the second inequality above. Since w must be between 356 and 364, the first inequality above is the only reasonable choice.

Solve the inequality by adding 3 to both sides to get x < 12.  Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.

3w/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.

Absolute value problems always have two sides: one positive and one negative.

First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.

Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).

We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.