Start by isolating the absolute value.

Subtract seven on both sides.

$\displaystyle 3\left | x-1\right |+7 -7=6-7$

$\displaystyle 3\left | x-1\right |=-1$

Divide by three on both sides.

$\displaystyle \frac{3\left | x-1\right |}{3}=\frac{-1}{3}$

$\displaystyle \left | x-1\right |=-\frac{1}{3}$

There is no value of $\displaystyle x$ in an absolute value that will give a negative value since the absolute value will convert all numbers to positive.

The answer is:  $\displaystyle \textup{No solution.}$

What is inside the absolute value brackets could be positive or negative.

Positive: $\displaystyle 3-x = 19$

$\displaystyle -x = 16$

$\displaystyle x = -16$

Negative: $\displaystyle 3-x = -19$

$\displaystyle -x = -22$

$\displaystyle x= 22$

Separate the absolute value and solve both the positive and negative components of the absolute value.

$\displaystyle 8-2x = 3x-2$

$\displaystyle -(8-2x) = 3x-2$

Solve the first equation.  Add $\displaystyle 2x$ on both sides.

$\displaystyle 8-2x +2x= 3x-2+2x$

$\displaystyle 8= 5x-2$

Add two on both sides.

$\displaystyle 10=5x$

Divide by five on both sides.

$\displaystyle \frac{10}{5}=\frac{5x}{5}$

One of the solutions is $\displaystyle x=2$ after substitution is valid.

Evaluate the second equation.  Distribute the negative sign in the binomial.

$\displaystyle -8+2x= 3x-2$

Subtract $\displaystyle 2x$ on both sides.

$\displaystyle -8+2x-2x= 3x-2-2x$

$\displaystyle -8 = x-2$

Add two on both sides.

$\displaystyle -8+2 = x-2+2$

$\displaystyle x=-6$

If we substitute this value back to the original equation, the equation becomes invalid.  

The answer is:  $\displaystyle 2$

Divide by two on both sides to isolate the absolute value.

$\displaystyle \frac{2\left | 3x-3\right | }{2}=\frac{16}{2}$

The equation becomes:

$\displaystyle \left | 3x-3\right | = 8$

Split this equation into its positive and negative components.

$\displaystyle 3x-3 = 8$

$\displaystyle -(3x-3)=8$

Solve the first equation.  Add three on both sides.

$\displaystyle 3x-3+3 = 8+3$

$\displaystyle 3x=11$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{11}{3}$

The first solution is:  $\displaystyle \frac{11}{3}$

Evaluate the second equation.  Divide by negative one on both sides.

$\displaystyle \frac{-(3x-3)}{-1}=\frac{8}{-1}$

$\displaystyle 3x-3=-8$

Add three on both sides.

$\displaystyle 3x-3+3=-8+3$

$\displaystyle 3x=-5$

Divide by three on both sides.

$\displaystyle \frac{3x}{3}=\frac{-5}{3}$

The second solution is:  $\displaystyle -\frac{5}{3}$

The answers are:  $\displaystyle -\frac{5}{3}, \frac{11}{3}$

Isolate the absolute value by dividing both sides by negative three.

$\displaystyle \frac{-3\left | 9x+2\right | }{-3}=\frac{ 33}{-3}$

The equation becomes:  

$\displaystyle \left | 9x+2\right | = -11$

Recall that an absolute value cannot be negative since the absolute value will change negative values to positive.

The answer is:  $\displaystyle \textup{No solution.}$

Subtract $\displaystyle 3x$ on both sides.

$\displaystyle \left | -3x+2\right |+3x -3x= 6-3x$

$\displaystyle \left | -3x+2\right | = -3x+6$

Evaluate the positive and negative components of the absolute value.  The two equations are:

$\displaystyle -3x+2 = -3x+6$

$\displaystyle -(-3x+2 )= -3x+6$

Evaluate the first equation.  Add $\displaystyle 3x$ on both sides.

$\displaystyle -3x+2 +3x= -3x+6+3x$

$\displaystyle 2=6$

There is no solution for the first equation.

Evaluate the second equation.  Simplify the left side by distributing negative one through the binomial.

$\displaystyle 3x-2 = -3x+6$

Add $\displaystyle 3x$ on both sides.

$\displaystyle 3x-2 +3x= -3x+6+3x$

The equation becomes:

$\displaystyle 6x-2 = 6$

Add two on both sides.

$\displaystyle 6x-2 +2= 6+2$

$\displaystyle 6x=8$

Divide by six on both sides and reduce the fraction.  Verify that the answer will satisfy the original equation.

$\displaystyle \frac{6x}{6}=\frac{8}{6} = \frac{4}{3}$

The answer is:  $\displaystyle \frac{4}{3}$

Evaluate by adding six on both sides.

$\displaystyle -2\left |\frac{1}{3}x-7 \right |-6+6 = -5+6$

$\displaystyle -2\left |\frac{1}{3}x-7 \right |=1$

Divide by negative two on both sides.

$\displaystyle \frac{-2\left |\frac{1}{3}x-7 \right |}{-2}=\frac{1}{-2}$

The equation becomes:  $\displaystyle \left |\frac{1}{3}x-7 \right | = -\frac{1}{2}$

Recall that absolute values will convert all negative values to positive.  No matter what $\displaystyle x$ would evaluate to be, it will never become negative half.

The answer is:   $\displaystyle \textup{No solution.}$

Divide both sides by eight.

$\displaystyle \frac{8\left |x-3 \right | }{8}= \frac{7}{8}$

$\displaystyle \left |x-3 \right | = \frac{7}{8}$

Split the absolute value into its positive and negative components.

$\displaystyle x-3 = \frac{7}{8}$

$\displaystyle -(x-3 )= \frac{7}{8}$

Solve the first equation.  Multiply both sides by eight to eliminate the fraction.

$\displaystyle 8(x-3) = \frac{7}{8}\cdot 8$

$\displaystyle 8(x-3) = 7$

Use distribution to simplify.

$\displaystyle 8x-24 = 7$

Add 24 on both sides, and then divide both sides by eight.

$\displaystyle 8x-24 +24= 7+24$

$\displaystyle 8x=31$

$\displaystyle \frac{8x}{8}=\frac{31}{8}$

The first solution is $\displaystyle \frac{31}{8}$.

Solve the second equation.  Divide by negative one on both sides.

$\displaystyle \frac{-(x-3 )}{-1}= \frac{\frac{7}{8}}{-1}$

The equation becomes:

$\displaystyle x-3 = -\frac{7}{8}$

Add three on both sides.  This is the same as adding $\displaystyle \frac{24}{8}$ on both sides.

$\displaystyle x-3+(3) = -\frac{7}{8}+(\frac{24}{8})$

$\displaystyle x=\frac{17}{8}$

The answers are:  $\displaystyle \frac{17}{8}, \frac{31}{8}$

Solve.

$\displaystyle 6-\left |2(x+4) \right |=0$

Step 1: Isolate the absolute value by subtracting $\displaystyle 6$ from both sides of the equation.

$\displaystyle -\left | 2(x+4)\right |=-6$

Step 2: Divide -1 from both sides of the equation in order to get rid of the negative sign in front of the absolute value.

$\displaystyle \left | 2(x+4)\right |=6$

Step 3: Because this is an inequality, this equation can be solved in two parts as shown below.

Note:  $\displaystyle \left | f(x)\right |=a$ can be written as $\displaystyle f(x)=-a$ or $\displaystyle f(x)=a$

$\displaystyle 2(x+4)=-6$  or  $\displaystyle 2(x+4)=6$

Step 4: Solve both parts.

$\displaystyle 2(x+4)=-6$

Distribute the $\displaystyle 2$.

$\displaystyle 2x+8=-6$

Subtract $\displaystyle 8$ from both sides of the equation.

$\displaystyle 2x=-14$

Divide both sides of the equation by $\displaystyle 2$.

$\displaystyle x=-7$

$\displaystyle 2(x+4)=6$

Distribute the $\displaystyle 2$.

$\displaystyle 2x+8=6$

Subtract $\displaystyle 8$ from both sides of the equation.

$\displaystyle 2x=-2$

Divide both sides of the equation by $\displaystyle 2$.

$\displaystyle x=-1$

Step 5: Combine both parts using "or".

$\displaystyle x=-7$ or $\displaystyle x=-1$

Solution: $\displaystyle x=-7$ or $\displaystyle x=-1$

Given: $\displaystyle |10x+7|=50$ 

When given an absolute value recognize there are often multiple solutions. The reason why is best exhibited in a simpler example:

Given $\displaystyle |x|=1$ solve for values $\displaystyle X$ of that make this statement true. When you taken an absolute value of something you always end up with the positive number so both $\displaystyle -1$ and $\displaystyle 1$ would make this statement true. The solutions can also be written as $\displaystyle ±1$.

In the case of the more complicated equation $\displaystyle |10x+7|=50$ for the same reason there are potentially two solutions, which are shown by $\displaystyle ±(10x+7)=50$ as an absolute value will always end up creating a positive result.

To simplify the absolute value we must look at each of these cases:

$\displaystyle 1)$ Let's start with the positive case:

$\displaystyle 10x+7=50$

Just like a normal equation with one unknown we will simplify it by isolating $\displaystyle x$ by itself. This is first done by subtracting $\displaystyle 7$ from both sides leaving:
$\displaystyle 10x=43$

Next $\displaystyle 10$ is divided from both sides leaving $\displaystyle x=43/10$, as your final solution.

To check this solution it must be substituted in the original absolute value for $\displaystyle x$ and if it's a correct answer you'll end up with a true statement, so

$\displaystyle |10x+7|=50$, $\displaystyle x=43/10$

$\displaystyle |10(43/10)+7|$$\displaystyle *43/10*10=43$

so this becomes:

$\displaystyle |43+7|=50$

and the absolute value of $\displaystyle 50$ is $\displaystyle 50$ so you end up with a true statement. Therefore $\displaystyle x=43/10$  is a valid solution

$\displaystyle 2)$ Next let's solve for the negative case:

$\displaystyle -(10x+7)=50$

Distribute the negative sign, which is just $\displaystyle -1$ to make calculations easier and you'll get:

$\displaystyle -10x-7=50$

Next $\displaystyle 7$ can be added to both sides, giving

$\displaystyle -10x=57$

dividing by $\displaystyle -10$ leaves:

$\displaystyle x=-57/10$

 Checking this solution is done just as you did for the previous solution obtained.

Given $\displaystyle |10x+7|=50$, $\displaystyle x=-57/10$

substitute $\displaystyle -57/10$  in for $\displaystyle x$

$\displaystyle |10(-57/10)+7|=50$

multiply $\displaystyle 10*(-57/10)$ gives $\displaystyle -57$

so you obtain $\displaystyle |-57+7|=50$

adding $\displaystyle -57+7=-50$

and the absolute value of $\displaystyle -50$ is $\displaystyle 50$ thereby making this also a valid solution, therefore the two valid solutions are  $\displaystyle x=43/10$ and $\displaystyle -57/10$