Question

How do you solve $3x=\dfrac{1}{27}$ ?

Answer 1

Hint: We are asked to find the solution of $3x=\dfrac{1}{27}$ .
Firstly, we learn what the solution of the equation means then we will learn what linear equation is in 1 variable term. We use a hit and trial method to find the value of ‘x’. in this method we put the value of ‘x’ 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 apply algebra. We subtract, add or, multiply terms to get to our final term and get our required solution. We will also learn that doing the questions using algebraic tools makes them easy.

Complete step by step solution:
We are given that, we have $3x=\dfrac{1}{27}$ . We are asked to find the value of ‘x’, 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 then the equation is satisfied.
Now we learn about the equation on 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 $3x=\dfrac{1}{27}$ also has just one variable ‘x’.
We have to find the value of ‘x’ 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 ‘x’ and take which one fits the solution correctly.
Let $x=1$ .
We put $x=1$ in $3x=\dfrac{1}{27}$. We get –
$\Rightarrow 3\times 1=\dfrac{1}{27}$
By simplifying, we get –
$\Rightarrow 3=\dfrac{1}{27}$ .
Which is not true.
So, $x=1$ is not the solution to our problem.
Now, Let $x=2$ .
We put $x=2$ in $3x=\dfrac{1}{27}$. We get –
$\Rightarrow 3\times 2=\dfrac{1}{27}$
By simplifying, we get –
$\Rightarrow 6=\dfrac{1}{27}$ .
Which is not true.
So, $x=2$ is not the solution to our problem.
As we can see that the distance between left side and right side increased when we moved from $x=1$ and $x=2$ , which means we are on wrong line.
Now, we let the value of ‘x’ in fraction to check our value.
We let $x=\dfrac{1}{3}$
We put $x=\dfrac{1}{3}$ in $3x=\dfrac{1}{27}$. We get –
$\Rightarrow 3\times \dfrac{1}{3}=\dfrac{1}{27}$
By simplifying, we get –
$\Rightarrow 1=\dfrac{1}{27}$ .
Which is not true.
So, $x=\dfrac{1}{3}$ is not the solution to our problem.
Distance decreased so it means we are moving on the right line.
We let $x=\dfrac{1}{9}$
We put $x=\dfrac{1}{9}$ in $3x=\dfrac{1}{27}$. We get –
$\Rightarrow 3\times \dfrac{1}{9}=\dfrac{1}{27}$
By simplifying, we get –
$\Rightarrow \dfrac{1}{3}=\dfrac{1}{27}$ .
Which is not true.
So, $x=\dfrac{1}{9}$ is not the solution to our problem.
We let $x=\dfrac{1}{81}$
We put $x=\dfrac{1}{81}$ in $3x=\dfrac{1}{27}$. We get –
$\Rightarrow 3\times \dfrac{1}{81}=\dfrac{1}{27}$
By simplifying, we get –
$\Rightarrow \dfrac{1}{27}=\dfrac{1}{27}$ .
So, $x=\dfrac{1}{81}$ is the solution to our problem.
In this method we need to check whether we are moving on the right path or not every time by checking the distance between the right side and the left side so this makes it little time consuming.
We have another way in which we use algebraic tools to solve the problem.
We have $3x=\dfrac{1}{27}$ .
We divide both the sides by ‘3’. So, we get –
$\Rightarrow \dfrac{3x}{3}=\left( \dfrac{1}{27} \right)\div 3$ .
By simplifying, we get –
$\Rightarrow x=\dfrac{1}{81}$ .
Hence, $x=\dfrac{1}{81}$ is the solution.

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 one added ‘6’ with 3 of x made it 9x, this is wrong, we cannot add constant and variable at once. Only the same variables are added to each other.
When we add the variable the only constant part is added or subtracted.
That is $2x+2x=4x$ error like doing it $2x+2x=4{{x}^{2}}$ may happen so be careful there.
Remember when we divide positive terms by negative values the solution we get is a negative term this may happen if we skip the (-) sign.
We need to choose the best techniques as per the question if the question is defined using variable only time so hit and trials would be ok but if the equation has more time variable then we go for the arithmetic tool.