Question

How do you solve 16=4n?

Answer 1

We are asked to find the solution of 16=4n. Firstly, we will learn what the solution of the equation means then we will learn what linear equation is in 1 variable term. We will use a hit and trial method to find the value of ‘n’. In this method we put the value of ‘n’ one by one by hitting arbitrary values and looking for needed values. Once we work with a hit and trial method we will try another method where we will apply algebra. We subtract, add or multiply terms to get to our final term and get our required solution. We will also learn to do the questions using algebraic operators to make them easy.

Complete answer:
We are given that 16=4n. We are asked to find the value of ‘n’, or we are asked how we will be able to solve this expression.
Solution of any problem is that value which when put into the given problem the equation is satisfied.
Now we learn about the equation that it is a one variable. One variable simply represents the equation that has one variable (say x, y, or z) and another one constant.
For Example:
x+2=4, 2-x=2, 2x, 2y etc.
Our equation 16=4n also has just one variable ‘n’.
We have to find the value of ‘n’ which will satisfy our given equation.
Firstly we try by the method of hit and trial. In which we will put a different value of ‘n’ and take which
one fits the solution correctly.
Let n=0
We put n=0 in 16=4n
We get, 16=0.
Which is not true so, n=0 is not the solution.
Now, let n=1.
We put n=1 in 16=4n.
We get $16=4\times 1$.
16=4
Which is not true so, n=1 is not the solution.
Now, let n=2.
We put n=2 in 16=4n.
We get $16=4\times 2=8$.
Which is not true so, n=2 is not the solution.
Now, let n=4.
We put n=4 in 16=4n.
We get $16=4\times 4=16$.
Which is true so, n=4 is the solution to our problem.
This method is totally a hit and miss method so we will use another method to solve the same problem.
In this method we will make use of algebraic operators like $+,-,\times ,\div$ to solve the problems.
We are given 16=4n.
Now we will divide both the sides by 4. So, we will get,
$4=n\left[ \because \dfrac{16}{4}=4\text{and}\dfrac{4n}{4}=n \right]$
Hence n=4 is the solution of our problem.

Note: Remember that, we cannot add the variable to the constant. Usual mistakes like this where one
adds constants with variables usually happen.
For example: 3x+6=9x, here if one adds ‘6’ with 3x and gets it as 9x, then it is wrong, since we cannot add constants and variables at once. Only the same variables can be added to each other. When we add the variable the only constant part is added or subtracted, the variable remains the same. That is we know that 2x+2x=4x but doing it as $2x+2x=4{{x}^{2}}$is wrong so we have to be careful. Remember, when we divide positive term by negative value the solution we get is a negative term but it can occur that we forget to put the negative sign in that term.