\(\displaystyle \left | 3x - 7 \right |=8\) really consists of two equations: \(\displaystyle 3x - 7 = \pm 8\)

We must solve them both to find two possible solutions.

\(\displaystyle 3x - 7 = 8 \Rightarrow 3x = 15\Rightarrow x = 5\)

\(\displaystyle 3x - 7 = - 8 \Rightarrow 3x = -1\Rightarrow x = -1/3\)

So \(\displaystyle x=5\) or \(\displaystyle x=\) \(\displaystyle \dpi{100} -\frac{1}{3}\).

It's actually easier to solve for the complement first.  Let's solve \(\displaystyle \left | 2x-5 \right |< 3\).  That gives -3 < 2x - 5 < 3.  Add 5 to get 2 < 2x < 8, and divide by 2 to get 1 < x < 4.  To find the real solution then, we take the opposites of the two inequality signs.  Then our answer becomes \(\displaystyle x\leq 1 \textsc{ or } x\geq 4\).

Set \(\displaystyle f (x) = 0\) and solve for \(\displaystyle x\):

\(\displaystyle f (x) = 0\)

\(\displaystyle \left | 4x + B \right | - 8 = 0\)

\(\displaystyle \left | 4x + B \right | = 8\)

Rewrite as a compound equation and solve each part separately:

\(\displaystyle 4x + B = -8 \textrm{ or } 4x + B = 8\)

\(\displaystyle 4x + B = -8\)

\(\displaystyle 4x + B -B = -B -8\)

\(\displaystyle 4x = -B -8\)

\(\displaystyle 4x \div 4 =\left ( -B -8 \right ) \div 4\)

\(\displaystyle x = \frac{-B -8 }{4}\)

\(\displaystyle 4x + B = 8\)

\(\displaystyle 4x + B -B = -B + 8\)

\(\displaystyle 4x = -B +8\)

\(\displaystyle 4x \div 4 =\left ( -B +8 \right ) \div 4\)

\(\displaystyle x = \frac{-B +8 }{4}\)

We can rewrite this as an equation, where \(\displaystyle N\) is the number in question:

\(\displaystyle N = |N| - 10\)

A nonnegative number is equal to its own absolute value, so if this number exists, it must be negative.

In thsi case, \(\displaystyle |N| = - N\), and we can rewrite that equation as

\(\displaystyle N = -N- 10\)

\(\displaystyle N +N= -N- 10 +N\)

\(\displaystyle 2N = -10\)

\(\displaystyle 2N \div 2 = -10\div 2\)

\(\displaystyle N = -5\)

This is the only number that fits the criterion.

Since \(\displaystyle -1< n< 0\), we know the following:  

\(\displaystyle 0< n^2< 1\) ;

\(\displaystyle -1< \frac{n}{2}< 0\);

\(\displaystyle \frac{1}{n^2}>1\);

\(\displaystyle \frac{1}{n}< -1\);

\(\displaystyle 0< n+1< 1\).

Also, we need to compare absolute values, so the greatest one must be either \(\displaystyle \left | \frac{1}{n^2}\right |\) or \(\displaystyle \left | \frac{1}{n}\right |\).

We also know that \(\displaystyle \left | n^2\right |< \left | n\right |\) when \(\displaystyle -1< n< 0\).

Thus, we know for sure that \(\displaystyle \left | \frac{1}{n^2}\right |>\left | \frac{1}{n}\right |\).

We can rewrite this as an equation, where \(\displaystyle N\) is the number in question:

\(\displaystyle N = 2\cdot |N| - 20\)

If \(\displaystyle N\) is nonnegative, then \(\displaystyle | N | = N\), and we can rewrite this as 

\(\displaystyle N = 2N - 20\)

Solve:

\(\displaystyle N -2N= 2N - 20-2N\)

\(\displaystyle -N= - 20\)

\(\displaystyle N = 20\)

If \(\displaystyle N\) is negative, then \(\displaystyle | N | =- N\), and we can rewrite this as 

\(\displaystyle N = 2\left (-N \right ) - 20\)

\(\displaystyle N = -2N - 20\)

\(\displaystyle N + 2N= -2N - 20 + 2N\)

\(\displaystyle 3N=- 20\)

\(\displaystyle 3N \div 3=- 20\div 3\)

\(\displaystyle N=- 6\frac{2}{3}\)

The numbers \(\displaystyle - 6\frac{2}{3}, 20\) have the given characteristics.

The correct answer is that there is no \(\displaystyle x\).

We start by adding \(\displaystyle 1\) to both sides giving

\(\displaystyle -\frac{5}{2} \left|x-5\right|=14\)

Then multiply both sides by \(\displaystyle 2\).

\(\displaystyle -5\left|x-5|=28\)

Then divide both sides by \(\displaystyle -5\)

\(\displaystyle |x-5|=-\frac{28}{5}\)

Now it is impossible to go any further. The absolute value of any quantity is always positive (or sometimes \(\displaystyle 0\)). Here we have the absolute value of something equaling a negative number. That's never possible, hence there is no \(\displaystyle x\) that makes this a true equation.

We start by isolating the expression with the absolute value:

\(\displaystyle 32-\left | x-23\right |=13\)  becomes \(\displaystyle -\left | x-23\right |=13-32=-19\)

So: \(\displaystyle x-23=19\) or \(\displaystyle x-23=-19\)

We then solve the two equations above, which gives us 42 and 4 respectively.

So the solution is \(\displaystyle x={42,4}\)

We proceed as follows

\(\displaystyle |2x-5| +3 =5\) (Start)

\(\displaystyle |2x -5| =2\) (Subtract 3 from both sides)

\(\displaystyle 2x-5 =2\) or \(\displaystyle 2x-5 = -2\) (Quantity inside the absolute value can be positive or negative)

\(\displaystyle 2x = 7\) or \(\displaystyle 2x = 3\) (add five to both sides)

\(\displaystyle x = \frac{7}{2}\) or \(\displaystyle x = \frac{3}{2}\)

Another way to say this is \(\displaystyle x = \frac{7}{2},\frac{3}{2}\)

To solve an inequality we need to remember what the absolute value sign says about our expression. In this case it says that

     \(\displaystyle \left | 2m-\frac{1}{2}\right |< 4\)

can be written as

\(\displaystyle 2m-\frac{1}{2}< 4\) Of \(\displaystyle 2m-\frac{1}{2}>-4\).

Rewriting this in one inequality we get:

\(\displaystyle -4< 2m-\frac{1}{2}< 4\)

From here we add one half to both sides .

\(\displaystyle -\frac{7}{2}< 2m< \frac{9}{2}\)

Finally, we divide by two to isolate and solve for m.

\(\displaystyle -\frac{7}{4}< m< \frac{9}{4}\)

\(\displaystyle -1.75< m< 2.25\)

Only \(\displaystyle \frac{7}{4}\) is between -1.75 and 2.25