Question

How do you solve $ 6x+1=4 $?

Answer 1

Hint: We are asked to find the solution of $ 6x+1=4 $. First, we will learn what a linear equation is in one variable. Then to find the solution of $ 6x+1=4 $ we will first use the method of hit and trial, we will put the value of x and look if it satisfying the condition, another method we will try is, use of algebraic operation like addition, subtraction, multiplication, and division to get our answer quickly and easily.

Complete step by step answer:
We are given the equation as $ 6x+1=4 $. We are asked to find the value of x or we can say we are asked how we are able to solve this expression.
We will learn about, equations in one variable to simply represent the equation that has one variable (say x, y, or z) and others are 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 which we will put the value of x and look if it hits the correct spot (solution) or not.
We have $ 6x+1=4 $ we have to find the value of x for which it is true.
We put x = 0, we get $ 6x+1=6\times 0+1=1\ne 4 $ .
So x = 0 is not the solution to one problem.
If we put x = -1 we get $ 6x+1=6\times \left( -1 \right)+1=-6+1=-5\ne 4 $ .
Again x = -1 is also not the solution.
If we put $ x=\dfrac{-5}{6} $ we get $ 6x+1=6\times \left( \dfrac{-5}{6} \right)+1=-5+1=-4\ne 4 $ .
So $ x=\dfrac{-5}{6} $ is also not the solution.
If we take $ x=\dfrac{1}{2} $ we get $ 6x+1=6\times \left( \dfrac{1}{2} \right)+1=3+1=4 $ .
So $ x=\dfrac{1}{2} $ this is the required solution.
Another method is the use of algebraic operations.
We have $ 6x+1=4 $ ,
We subtract 1 from both sides and we get $ 6x+1-1=4-1 $.
As -1+1 = 0 and 4-1 = 3 so we get 6x = 3.
Now divide both sides by 6 we get $ \dfrac{6x}{6}=\dfrac{3}{6} $ .
Simplifying we get $ x=\dfrac{1}{2} $ .
So $ x=\dfrac{1}{2} $ is the required solution.

Note:
While solving this problem, hit and the trial will become more lengthy as sometimes we may start from a point and move along say positive direction but our solution may lie into the negative side, so we will keep finding and still get nothing, so we use an algebraic method in which we cancel all the terms by addition, subtraction, multiplication, and division this will make solution easy and short.