Equations with Absolute Values: Solving the problem

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 $x=-7$.

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 $x=-6$.

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 $\left|-2x\right|=16$, 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 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 $x = 15$.

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 $\left| y - 3 \right| = 7$, 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 $\left| a - 2 \right| = 6$, 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 this exercise, we need to note that the left side is in absolute value.

Absolute value checks the distance of a number from zero, meaning its solution is always positive.

Therefore, we have two possibilities, either the numbers inside will be positive or negative,

In other words, we check two options, in one what's inside the absolute value is positive and in the second it's negative.

6x-12=6

6x=18

x=3

This is the first solution

-(6x-12)=6
-6x+12=6
-6x=6-12
-6x=-6
6x=6
x=1

And this is the second solution,

So we found two solutions,

x=1, x=3

And that's the solution!

To solve $\left| 2x + 1 \right| = 9$, we consider two cases:

1. $2x + 1 = 9$ implies $2x = 8 \to x = 4$

2. $2x + 1 = -9$ implies $2x = -10 \to x = -5$

Thus, the solutions are $x = 4$ and $x = -5$.

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 $x = -1$.

To solve $3|x + 2| = 18$, first divide by 3 to get $|x + 2| = 6$.

This absolute value equation means $x + 2 = 6$ or $x + 2 = -6$. Solving these:

1. $x + 2 = 6$ leads to $x = 4$.

2. $x + 2 = -6$ leads to $x = -8$.

Therefore, the solutions are $x = 4$ and $x = -8$.

To solve $\left|2x - 4\right|=8$, we split it into two cases based on the nature of absolute values:

1) $2x - 4=8$:

Add $4$ to both sides: $2x=12$.

Divide both sides by $2$ to get $x=6$.

2) $2x - 4=-8$:

Add $4$ to both sides: $2x=-4$.

Divide both sides by $2$ to get $x=-2$.

So, the solutions are $x=-2$ and $x=6$.

To solve $\left|4x+1\right|=9$, split into two scenarios:

1) $4x+1=9$:

Subtract $1$ from both sides to get $4x=8$.

Divide both sides by $4$ to find $x=2$.

2) $4x+1=-9$:

Subtract $1$ from both sides to get $4x=-10$.

Divide both sides by $4$ to find $x=-2.5$.

Thus, $x=2$ and $x=-2.5$ are solutions.

To solve $\left|-5x + 2\right|=13$, address the two absolute value cases:

1) $-5x + 2=13$:

Subtract $2$ from both sides to get $-5x=11$.

Divide by $-5$ to find $x=-2.2$.

2) $-5x + 2=-13$:

Subtract $2$ from both sides to get $-5x=-15$.

Divide by $-5$ for $x=3$.

The solutions are $x=3$ and $x=-2.2$.