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 \(\displaystyle -57/10\)

Solve for \(\displaystyle x\) given \(\displaystyle |7x+14+22x|-72=-3x\) 

When given an absolute value equation 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 \(\displaystyle |7x+14+22x|-72=-3x\), however before proceeding let's simplify this equation a little as there are two \(\displaystyle x\) terms that can be added together within the absolute value. These are \(\displaystyle 7x\) and \(\displaystyle 22x\).

When added together this gives \(\displaystyle 29x\),

thereby giving you \(\displaystyle |29x+14|-72=-3x\)

For the same reason as shown in the case of \(\displaystyle |x|=1\)  there are potentially two solutions to this equation, which are shown by \(\displaystyle ±(29x+14)-72=-3x\) as an absolute value will always end up creating a positive result. Make sure you apply the parentheses to only the portion of the equation with the absolute value otherwise your answer will be incorrect.

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

 Let's start with the positive case:

\(\displaystyle (29x+14)-72=-3x\)

Just like a normal equation with one unknown we will simplify it by isolating \(\displaystyle x\) by itself. This is first done by combining like terms and getting \(\displaystyle x\) on one side of the equation.

We can first subtract \(\displaystyle 72\) from \(\displaystyle 14\) which is \(\displaystyle -58\), this gives you:

\(\displaystyle 29x-58=-3x\)

Next we can subtract \(\displaystyle 29x\) from both sides of the equation.

\(\displaystyle -58=-26x\)

Dividing both sides by \(\displaystyle -26\) gives you a final answer of \(\displaystyle x=58/26\) this however can be simplified to \(\displaystyle 29/16\)  as both the numerator and denominator are divisible by \(\displaystyle 2\).

 so  \(\displaystyle x=29/16\)

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

\(\displaystyle |29x+14|-72=-3x\),\(\displaystyle x=29/16\)

\(\displaystyle |29(29/16)+14|-72=-3(29/16)\)

Simplify the equation by multiplying \(\displaystyle 29\) by \(\displaystyle 29/16\) and \(\displaystyle -3\) by \(\displaystyle 29/16\)

\(\displaystyle 29*29/16=841/16\)

\(\displaystyle -3*29/16=-87/16\)

This leaves our equation with

\(\displaystyle |841/16+14|-72=-87/16\) 

Next add \(\displaystyle 72\) to both sides of the equation:

\(\displaystyle |841/16+14|=72-87/16\)

In order to simplify \(\displaystyle 72-87/16\)  you must find a common denominator, which is most easily done by multiplying \(\displaystyle 72*16/16\).

This leaves:

\(\displaystyle |841/16+14|=1152/16-87/16\)

\(\displaystyle |841/16+14|=1065/16\)

Similarly a common denominator is found for \(\displaystyle 841/16\)  and \(\displaystyle 14\) by multiplying \(\displaystyle 14\) by  \(\displaystyle 16/16\) which gives you \(\displaystyle 224/16\)

this leaves:

\(\displaystyle |841/16+14*16/16|=1065/16\)

\(\displaystyle |841/16+224/16|=1065/6\)

simplifying further gives you 

\(\displaystyle |1065/6|=1065/6\) which is a true statement, so \(\displaystyle x=29/16\) is a valid solution

Next let's solve for the negative case:

\(\displaystyle -(29x+14)-72=-3x\)

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

\(\displaystyle -29x-14-72=-3x\) 

Combine like terms:

\(\displaystyle -86=26x\)

\(\displaystyle x=-86/26\)

This can be simplified to \(\displaystyle -43/13\) as both the numerator and denominator are divisible by \(\displaystyle 2\), therefore you final answer is \(\displaystyle x=-43/13\)

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

Given \(\displaystyle |29x+14|-72=-3x,\)\(\displaystyle x=-43/13\)

substitute \(\displaystyle -43/13\) in for \(\displaystyle x\)

\(\displaystyle |29(-43/13)+14|-72=-3(-43/13)\)

Multiplying \(\displaystyle 29\) by \(\displaystyle (-43/13)\) gives \(\displaystyle -1247/13\)

Multiplying \(\displaystyle -3*(-43/13)\) gives \(\displaystyle 129/13\)

So the equation simplifies to:

\(\displaystyle |-1247/13+14|-72=129/13\)

Next \(\displaystyle 72\) can be added to both sides of the equation giving:

\(\displaystyle |-1247/13+14|=129/13+72\) 

Now common denominators must be found for \(\displaystyle 129/13+72\) and \(\displaystyle -1247/13+14.\)

The common denominator for \(\displaystyle 129/13\) is found by multiplying \(\displaystyle 72\) by \(\displaystyle 13/13\). This gives you \(\displaystyle 936/13+129/13\). This can then be simplified through addition and gives \(\displaystyle 1065/13\).

The common denominator of \(\displaystyle -1247/13+14\) is found by multiplying \(\displaystyle -1247/13*14/14\) and \(\displaystyle 14*(13*14)/(13*14)\)

\(\displaystyle -1247/13*14/14=-17458/182\)

\(\displaystyle 14*(13*14)/(13*14)=14*182/182=2548/182\)

so\(\displaystyle -17458/182+2548/182=-14910/182\) 

The simplified equation becomes

\(\displaystyle |-19410/182|=1065/13\)

Through dividing \(\displaystyle -19410/182\) by \(\displaystyle 13\) you get \(\displaystyle -1065/13\) and the absolute value of \(\displaystyle -19410/182\)  is \(\displaystyle 1065/13\), so you get \(\displaystyle 1065/13\). This is true statement

so \(\displaystyle x=-43/13\) is also a valid solution.

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

\(\displaystyle \left | 2x-5\right |=3x\)

\(\displaystyle 2x-5=3x\) 

and 

\(\displaystyle 2x-5=-(3x)\)

This gives us:

\(\displaystyle -5=x\) and \(\displaystyle 5x=5\)

\(\displaystyle x=-5\ and\ 1\)

However, this question has an \(\displaystyle x\) outside of the absolute value expression, in this case \(\displaystyle 3x\). Thus, any negative value of \(\displaystyle x\) will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, \(\displaystyle x=-5\) is an extraneous solution, as \(\displaystyle \left | 2x-5\right |\) cannot equal a negative number.

Our final solution is then

\(\displaystyle x=1\)

\(\displaystyle \small \left | x-4 \right | \geq 7\)

\(\displaystyle \small x-4\geq7\ or\ x-4\leq-7\)

\(\displaystyle \small x\geq11\ or\ x\leq-3\)