How do you solve the algebraic equation \[8z = 4\left( {2z + 1} \right)\]?

Hint:Given a linear equation in one variable which is varying in \[x\]. We know that the standard form of linear equation in one variable is given by \[ax + b = 0\], where the value of the coefficient of the x term is not equal to zero. That is \[a \ne 0\], and \[b\] may or may not be zero. Here we find the solution of the equation by rearranging the like terms together.

Complete answer:
Considering the given linear equation \[8z = 4\left( {2z + 1} \right)\] below:
\[ \Rightarrow 8z = 4\left( {2z + 1} \right)\]
Now, simplify the right hand side of the equation by multiplying 4 to the elements of parenthesis as shown below.
\[ \Rightarrow 8z = 8z + 4\]
Now converting the above given linear equation in one variable into the standard form of the linear equation in one variable, as shown below:
By transferring the like terms and the unlike terms on one side of the above equation, as shown below:
\[ \Rightarrow 8z - 8z - 4 = 0\]
Now simplifying the above equation as shown below:
\[ \Rightarrow 0z - 4 = 0\]
Simplify the equation further, it is observed that the variable z is eliminated from the equation as shown below.
\[ \Rightarrow - 4 = 0\]
It is observed that the resultant equation is not true as we know the negative of four is not equal to zero.
Thus, the given equation cannot be satisfied by any value of variable z.
Hence, there is no solution of the value of z for this given equation \[8z = 4\left( {2z + 1} \right)\].

Note:
Please note that instead of grouping like terms and the unlike terms together we can actually group all the terms on one side of the linear equation which is the similar form of the standard linear equation \[ax + b = 0\], and then divide the equation with \[a\], and then moving the constant to the other side which brings the solution to \[x = - \dfrac{b}{a}\].