How do you solve $5(x-4)=5x+12$ ?

Hint: In this question, we have to find the value of x. Since, the given equation is linear, because it consists of only one variable and many constants. Thus, to solve this problem, we will first apply the distributive property and then some basic mathematical rules. So, we first add 4 on both sides of the equation and then make the necessary calculations. Then, we will subtract 5x on both sides of the equation and combine the like terms. After that, we will divide 3 on both sides of the equation and make the necessary calculations. In the last, we multiply both sides by (-1), to get the value of x, which is our required answer.

Complete step by step solution:
According to the problem, we have to find the value of x.
So, we will use distributive property to get the solution.
The equation given to us is $5(x-4)=5x+12$ ------------- (1)
Let us first apply the distributive property $a(b-c)=ab-ac$ on the left-hand side in equation (1), we get
$5(x)-5(4)=5x+12$
O further simplification, we get
$5x-20=5x+12$
Firstly, we will add 20 on both sides in the equation (1), we get
$5x-20+20=5x+12-20$
Now, the same terms with opposite signs cancel out each other on the left-hand of the above equation, therefore we get
$\Rightarrow 5x=5x-8$
Now, we will subtract $5x$ on both sides in the above equation, we get
$\Rightarrow 5x-5x=5x-8-5x$
As we know, the same terms with opposite signs cancel out each other on the right-hand of the above equation, thus we get
$\Rightarrow 0=-8$
Thus, we see that the variable x is canceled, therefore there is no value of x.
Therefore, for the equation $5(x-4)=5x+12$ , there is no value of x, which is our required answer.

Note:
Make all the necessary calculations properly to avoid mistakes. One of the alternative methods is after 5x = 5x - 8, subtract both sides of the equation by 21, then cancel out the same terms with opposite signs. After that, again subtract 5x on both sides of the equation and we get no solution for the equation, which is our required answer.