Absolute Value: Additional Practice (Set 1)

Add inside the bars first. Two negatives combine further left: $-3$ and $-9$ give $-12$.

$\lvert (-3) + (-9) \rvert = \lvert -12 \rvert = 12$

An integer six units from zero can sit on either side of zero.

$\lvert 6 \rvert = 6 \quad\text{and}\quad \lvert -6 \rvert = 6$

Those are the two integers, $6$ and $-6$.

The bars group the product, so multiply inside first. A positive times a negative is negative, giving $-12$, and then the bars measure its distance from zero.

$\lvert 4 \times (-3) \rvert = \lvert -12 \rvert = 12$

Each pair of bars is its own group, so take each absolute value, then add.

$\lvert -2 \rvert + \lvert -3 \rvert + \lvert -4 \rvert = 2 + 3 + 4 = 9$

Take the absolute value of each, since the question asks about distance from zero.

$\lvert -14 \rvert = 14, \quad \lvert 11 \rvert = 11, \quad \lvert -9 \rvert = 9, \quad \lvert 13 \rvert = 13$

The greatest distance is $14$, from $-14$.

Start at zero and go no further than two units in either direction. List what you land on: $-2, -1, 0, 1, 2$.

$\lvert -2 \rvert = 2, \; \lvert -1 \rvert = 1, \; \lvert 0 \rvert = 0, \; \lvert 1 \rvert = 1, \; \lvert 2 \rvert = 2$

That is five integers.

The bars group the quotient, so divide inside first. A negative divided by a positive is negative, giving $-4$, and then the bars measure its distance from zero.

$\lvert -20 \div 5 \rvert = \lvert -4 \rvert = 4$

A number and its opposite share one absolute value, because both are the same distance from zero.

$\lvert -8 \rvert = 8 = \lvert 8 \rvert$

So the third statement is true. The first gives $16$, the second a negative, and the fourth a false inequality.

Evaluate each group. The first is $\lvert -5 \rvert = 5$, and the second is $\lvert 2 - 9 \rvert = \lvert -7 \rvert = 7$.

$5 - 7 = -2$

The minus between the two groups keeps its meaning, so the result can be negative.

Evaluate inside each pair of bars first. The first is $\lvert (-4) + 4 \rvert = \lvert 0 \rvert = 0$, and the second is $\lvert -6 - 2 \rvert = \lvert -8 \rvert = 8$.

$0 + 8 = 8$

Take the absolute value of each: $\lvert -9 \rvert = 9$, $\lvert 3 \rvert = 3$, $\lvert -5 \rvert = 5$, $\lvert 8 \rvert = 8$.

$9 > 8 > 5 > 3$

Ordering the original numbers by those distances, greatest first, gives $-9,\; 8,\; -5,\; 3$.

Two integers are ten units from zero: $10$ and $-10$. The condition that $x$ is positive selects one.

$\lvert 10 \rvert = 10 \text{ and } 10 > 0$

So $x = 10$. The sign condition removes the ambiguity that $\lvert x \rvert = 10$ alone would leave.