Recall that with absolute values and "less than" inequalities, we have to hold the following:

12x + 3y < 15

AND

12x + 3y > –15

Otherwise written, this is:

–15 < 12x + 3y < 15

In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:

–15 – 12x < 3y < 15 – 12x

Now, we have to divide each element by 3:

(–15 – 12x)/3 < y < (15 – 12x)/3

This simplifies to:

–5 – 4x < y < 5 – 4x

This inequality could be rewritten as:

4x + 14 > 30  OR 4x + 14 < –30

Solve each for x:

4x + 14 > 30; 4x > 16; x > 4

4x + 14 < –30; 4x < –44; x < –11

Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.

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$\displaystyle -2x\leq 5$. 

Divide both sides by 2 (using division rule) and you will get $\displaystyle -x\leq \frac{5}{2}$ which is the same as $\displaystyle x\geq \frac{5}{2}$

Subtract 5 from both sides of the inequality:

$\displaystyle \frac{a}{5}> 1$

Multiply both sides by 5:

$\displaystyle a> 5$

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 $\displaystyle x$ is on a side by itself:

$\displaystyle x+4>2x-2$

$\displaystyle x+4-x>2x-2-x$

$\displaystyle 4>x-2$

$\displaystyle 4+2>x-2+2$

$\displaystyle 6>x$

For this equation, $\displaystyle x$ 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.

$\displaystyle \dpi{100} \small -2y+7>-7+y$Subtract $\displaystyle y$ from both sides:

$\displaystyle -2y+7-y>-7+y-y$

$\displaystyle -3y+7>-7$

Subtract 7 from both sides:

$\displaystyle -3y+7-7>-7-7$

$\displaystyle -3y>-14$

Divide both sides by $\displaystyle \dpi{100} \small -3$:

$\displaystyle \frac{-3y}{-3}>\frac{-14}{-3}$

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

$\displaystyle y< \frac{14}{3}$

Since $\displaystyle \dpi{100} \small y< \frac{14}{3}$ is not an answer, we must find an answer that, at the very least, does not contradict the fact that $\displaystyle y$ 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 $\displaystyle y$ 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:

$\displaystyle \\ |w -360| \leq 4 \\ |360 -360| \leq 4 \\0 \leq 4$

and:

$\displaystyle \\|w + 360| \geq 4 \\ |360 + 360| \geq 4\\ 720 \geq4$

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.