Question

How do you solve $ 3 = x + 3 - 5x $ ?

Answer 1

Hint: We can see that the given equation is a linear equation with one variable as the degree of the equation is $ 1 $ and only one variable is present in the equation. When we are given such equation to solve, it means we have to find the value of $ x $ by shifting all the terms without the variable $ x $ on the RHS, while keeping all the terms with the variable $ x $ on the LHS and then equating them will give us the value of $ x $ .

Complete step-by-step answer:
(i)
The equation given to us is:
 $ 3 = x + 3 - 5x $
Since it has only one variable present in it and the degree of the equation is $ 1 $ as the highest power of the variable $ x $ is $ 1 $ , we can say that it is a linear equation with one variable. Solving an equation with one variable is the same as calculating the value of the variable i.e., $ x $ .
Therefore, to find the value of $ x $ , we will first write the equation like this:
 $ x + 3 - 5x = 3 $
(ii)
Now, we have to shift all the terms containing the variable $ x $ to LHS and the rest of the terms to RHS. Since, all the terms with $ x $ are already in LHS, we will just shift $ + 3 $ to RHS. So, after shifting, we will get:
 $ x - 5x = 3 - 3 $
(iii)
Now, we will subtract $ 5x $ from $ x $ and $ 3 $ from $ 3 $ as shown in the equation:
 $ - 4x = 0 $
(iv)
As we have to find the value of $ x $ , we will divide both the sides by $ - 4 $ . Therefore, we will get:
 $ \dfrac{{ - 4x}}{{ - 4}} = \dfrac{0}{{ - 4}} $
Simplifying it further, we will get:
 $ x = 0 $
Hence, the solution of $ 3 = x + 3 - 5x $ is $ x = 0 $ .
So, the correct answer is “0”.

Note: Always remember that in questions like this, our approach should be keeping the variable $ x $ in LHS, and rest everything on RHS to isolate $ x $ because this is how we will find the value of $ x $ . Also, note that the sign of the term changes when we shift that term from one side to the other. For example, when we shifted $ + 3 $ from LHS to RHS in second step, it changed it sign and became $ - 3 $