Question

How do you solve $ \dfrac{1}{3}+x=\dfrac{5}{6} $ ?

Answer 1

Hint: We are asked to find the solution of $ \dfrac{1}{3}+x=\dfrac{5}{6} $ , firstly we learn that what is a linear equation in 1 variable term.
We use the hit and trial method to find the value of ‘x’ such that $ \dfrac{1}{3}+x=\dfrac{5}{6} $.
In this method, we put the value of ‘x’ one by one by hitting arbitrary value and look for the needed value. Another method is to apply algebra. We subtract the term to get to our final term and get our required solution.

Complete step by step answer:
We are given that we have $ \dfrac{1}{3}+x=\dfrac{5}{6} $ .
We are asked to find the value of ‘x’ or we say we are asked how are be able to solve this expression.
We learn about equations in one variable. The equation in one variable simple represents the equation that has one variable (say x, y, or z) and other one is constant.
For example:
 $ x+2=4,2-x=2,2x,2y $ etc.
Our equation $ x+3=5 $ also has just one variable ‘x’.
We have to find the value of ‘x’ which will satisfy our given equation.
We can go by the method of hit and trial in this method we try to get the value of solution by guessing a number and then try it in our given equation.
Let $ x=0 $ , in $ \dfrac{1}{3}+x=\dfrac{5}{6} $
We put $ x=0 $ , we get –
 $ \dfrac{1}{3}+0=\dfrac{5}{6} $
 $ \dfrac{1}{3}=\dfrac{5}{6} $ Not true.
So $ x=0 $ is not the solution.
We put $ x=1 $ , we get –
 $ \dfrac{1}{3}+1=\dfrac{5}{6} $
By solving we get –
 $ \dfrac{1+3}{3}=\dfrac{5}{6} $
 $ \dfrac{4}{3}=\dfrac{5}{6} $ , which is not true.
So, $ x=1 $ is also not solution.
We try another value $ x=\dfrac{1}{3} $ , we get –
 $ \dfrac{1}{3}+\dfrac{1}{3}=\dfrac{5}{6} $
 $ \dfrac{2}{3}=\dfrac{5}{6} $ which is not true.
So, $ x=\dfrac{1}{3} $ is also not true.
Now we put $ x=\dfrac{1}{2} $ in $ \dfrac{1}{3}+x=\dfrac{5}{6} $ we get –
 $ \dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6} $
 $ \dfrac{2+3}{6}+=\dfrac{5}{6} $
 $ \dfrac{5}{6}=\dfrac{5}{6} $
LHS = RHS
So, $ x=\dfrac{1}{2} $ is our solution.
This method is quite lengthy sometimes.
We also use the algebraic operation to solve such problems in an easy way.
We have –
 $ \dfrac{1}{3}+x=\dfrac{5}{6} $
Now, we subtract $ \dfrac{1}{3} $ on both side, we get –
 $ \dfrac{1}{3}+x-\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{3} $
Simplifying, we get –
 $ x=\dfrac{5}{6}-\dfrac{1}{3} $
Solving both hand side, we get –
 $ x=\dfrac{5-2}{6}=\dfrac{3}{6}=\dfrac{1}{2} $
So, we get $ x=\dfrac{1}{2} $ in the current solution.

Note:
While solving this problem hit and trial will become lengthier as sometimes we may start from a point and move in say positive direction but our solution lies on negative side. So, we will keep finding and still get nothing so we use algebra, in which we cancel all the terms by addition, subtraction, multiplication, and division. This will make the solution easy and short.