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 \(\displaystyle -57/10\)