Okay so I did the math.
1. Two solutions were found :<span><span>
z=-1
</span><span>z=2/3
</span></span><span>Step 1 :</span>Rearrange this Absolute Value Equation
Absolute value equalitiy entered
|2z-3| = 4z-1
<span>Step 2 :</span>Clear the Absolute Value Bars
Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.
The Absolute Value term is<span> |2z-3|
</span>For the Negative case we'll use -(2z-3)
For the Positive case we'll use (2z-3)
<span>Step 3 :</span>Solve the Negative Case
-(2z-3) = 4z-1
Multiply
-2z+3 = 4z-1
Rearrange and Add up
-6z = -4
Divide both sides by 6
-z = -(2/3)
Multiply both sides by (-1)
z = (2/3)
Which is the solution for the Negative Case
<span>Step 4 :</span>Solve the Positive Case
(2z-3) = 4z-1
Rearrange and Add up
-2z = 2
Divide both sides by 2
-z = 1
Multiply both sides by (-1)
z = -1
Which is the solution for the Positive Case
Step 5 :Wrap up the solution
<span> z=2/3
</span><span> z=-1
2. </span>-6 < x < 26/3
<span>Step 1 :</span>Rearrange this Absolute Value Inequality
Absolute value inequalitiy entered
|3x-4|+5 < 27
Another term is moved / added to the right hand side.
|3x-4| < 22
<span>Step 2 :</span>Clear the Absolute Value Bars
Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.
The Absolute Value term is<span> |3x-4|
</span>For the Negative case we'll use -(3x-4)
For the Positive case we'll use (3x-4)
<span>Step 3 :</span>Solve the Negative Case
-(3x-4) < 22
Multiply
-3x+4 < 22
Rearrange and Add up
-3x < 18
Divide both sides by 3
-x < 6
Multiply both sides by (-1)
Remember to flip the inequality sign
x > -6
Which is the solution for the Negative Case
<span>Step 4 :</span>Solve the Positive Case
(3x-4) < 22
Rearrange and Add up
3x < 26
Divide both sides by 3
x < (26/3)
Which is the solution for the Positive Case
<span>Step 5 :</span>Wrap up the solution
-6 < x < 26/3
Solution in Interval Notation
(-6,26/3)
HOPE THIS HELPS :D
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-8-%28-2%29%5D%5E2%2B%5B-3-5%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-8%2B2%29%5E2%2B%28-3-5%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B100%7D%5Cimplies%20d%3D10)
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-5%261%265%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-5%29%5D%5E2-%5By-1%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B5%29%5E2-%28y-1%29%5E2%3D25)
