Absolute Value and Inequality - Examples, Exercises and Solutions

To solve the equation $\left|-2x\right|=16$, we consider the definition of absolute value:

1. The expression inside the absolute value can be either positive or negative, but its absolute value is always positive.

2. Therefore, we set up two equations to solve:

$-2x = 16$ and $-2x = -16$

3. Solving the first equation:

$-2x = 16$

4. Divide both sides by -2:

$x = -8$

5. Solving the second equation:

$-2x = -16$

6. Divide both sides by -2:

$x = 8$

7. Therefore, the solution is:

$x=-8$ , $x=8$

To solve $\left|-3x\right|=15$, we consider both potential cases stemming from the absolute value:

1) $-3x=15$:

Divide both sides by $-3$ to get $x=-5$.

2) $-3x=-15$:

Divide both sides by $-3$ to get $x=5$.

Thus, the solutions are $x=-5$ and $x=5$.

To solve the equation $\left|3x\right|=21$, we consider the definition of absolute value:

1. The expression inside the absolute value can be either positive or negative, but its absolute value is always positive.

2. Therefore, we set up two equations to solve:

$3x = 21$ and $3x = -21$

3. Solving the first equation:

$3x = 21$

4. Divide both sides by 3:

$x = 7$

5. Solving the second equation:

$3x = -21$

6. Divide both sides by 3:

$x = -7$

7. Therefore, the solution is:

$x=-7$ , $x=7$

To solve the equation $\left|-4x\right|=12$, we consider the definition of absolute value:

1. The expression inside the absolute value can be either positive or negative, but its absolute value is always positive.

2. Therefore, we set up two equations to solve:

$-4x = 12$ and $-4x = -12$

3. Solving the first equation:

$-4x = 12$

4. Divide both sides by -4:

$x = -3$

5. Solving the second equation:

$-4x = -12$

6. Divide both sides by -4:

$x = 3$

7. Therefore, the solution is:

$x=-3$ , $x=3$

Let's solve the equation $-\left|x\right|=15$. Since the expression inside the absolute value can be either positive or negative, we consider two scenarios:

1. $-x = 15$: This implies $x = -15$.

2. $-(-x) = 15$: This simplifies to $x = 15$.

Thus, the solutions are $x = -15$ and $x = 15$.

The equation is $\left|x\right|=3$, which implies that $x$ can be 3 or -3. Hence, the solutions are $x=3$ and $x=-3$.

The equation is $\left|x\right|=4$. This means that the value of $x$ can be either 4 or -4. Thus, the solutions are $x=4$ and $x=-4$.

The equation is $\left|x\right|=6$, so $x$ could be 6 or -6. Therefore, the solutions are $x=6$ and $x=-6$.

The equation is $\left|x\right|=7$, which means that the absolute value of $x$ is 7. Therefore, $x$ could be 7 or -7. Thus, the solutions are $x=7$ and $x=-7$.

Let's solve the equation $-\left|x\right|=8$. The expression inside the absolute value can take two forms:

1. $-x = 8$: This gives $x = -8$.

2. $-(-x) = 8$: This simplifies to $x = 8$.

Therefore, the solutions are $x = -8$ and $x = 8$.

To solve $\left| a - 2 \right| = 6$, consider the cases:

1. $a - 2 = 6$ leads to $a = 8$

2. $a - 2 = -6$ leads to $a = -4$

Thus, the solutions are $a = 8$ and $a = -4$.

To solve the equation $\left|x+4\right|=10$, we split it into two separate equations:

1. $x+4=10$

2. $x+4=-10$

For the first equation:

$x+4=10$

Subtract 4 from both sides:

$x=6$

For the second equation:

$x+4=-10$

Subtract 4 from both sides:

$x=-14$

Thus, the solutions are $x=6$ and $x=-14$.

To solve $\left| y - 3 \right| = 7$, we need to consider the two possible cases for the absolute value equation.

1. $y - 3 = 7$ leads to $y = 10$

2. $y - 3 = -7$ leads to $y = -4$

Thus, the solutions are $y = 10$ and $y = -4$.

To solve $\left| z + 4 \right| = 12$, consider the two cases:

1. $z + 4 = 12$ gives $z = 8$

2. $z + 4 = -12$ gives $z = -16$

Thus, the solutions are $z = 8$ and $z = -16$.

To solve the equation $2|x - 4| = 10$, divide both sides by 2. This gives $|x - 4| = 5$.

The absolute value equation $|x - 4| = 5$ implies that $x - 4 = 5$ or $x - 4 = -5$. Solving these equations:

1. $x - 4 = 5$ gives $x = 9$.

2. $x - 4 = -5$ gives $x = -1$.

Thus, the solutions are $x = 9$ and $x = -1$.