How do you solve $ x+3=5 $?
Answer
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Hint: We are asked to find the solution of $ x+3=5 $ , firstly we learn that what is a linear equation in 1 variable then we use the hit and trial method to find the value of ‘x’ such that –
$ x+3=5 $, in this method we put the value of ‘x’ one by one by hitting arbitrary values and looking for needed values. Another method is to apply algebra to subtract terms to get to our final term and get our required solution.
Complete step by step answer:
We are given that $ x+3=5 $
We are asked to find the value of ‘x’ or we are asked how we will be able to solve this expression.
We learn about the equation in 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 $ 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
Like we put different values of ‘x’ and check which will satisfy our equation.
Like put x=0
We get $ x+3=0+3 $
By simplifying, we get –
$ x+3=3 $ which is not 5
So,
$ x+3\ne 5 $
X=0 is not our solution
We put x=1
We get –
$ \begin{align}
& x+3=1+3 \\
& =4 \\
\end{align} $
So, again
$ x+3=4\ne 5 $
So, x=1 is not our solution again.
We put x=2
We get –
$ \begin{align}
& x+3=2+3 \\
& =5 \\
\end{align} $
So, $ x+3=5 $
So, x=2 is our required solution.
This method is a bit extensive.
Another way is to go by algebra.
We have $ x+3=5 $
We subtract -3 from both sides, we get –
So, we get –
$ x+3-3=5-3 $
As $ 3-3=0 $ and $ 5-3=2 $
So, we get –
$ x=2 $
So, solution is $ x=2 $
We get –
For $ x+3=5 $
Solution is $ x=2 $
Note:
While solving this problem hit and the trial will become more lengthy as sometimes we may start from a point and more along say positive director but our solution lie into the 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.
$ x+3=5 $, in this method we put the value of ‘x’ one by one by hitting arbitrary values and looking for needed values. Another method is to apply algebra to subtract terms to get to our final term and get our required solution.
Complete step by step answer:
We are given that $ x+3=5 $
We are asked to find the value of ‘x’ or we are asked how we will be able to solve this expression.
We learn about the equation in 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 $ 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
Like we put different values of ‘x’ and check which will satisfy our equation.
Like put x=0
We get $ x+3=0+3 $
By simplifying, we get –
$ x+3=3 $ which is not 5
So,
$ x+3\ne 5 $
X=0 is not our solution
We put x=1
We get –
$ \begin{align}
& x+3=1+3 \\
& =4 \\
\end{align} $
So, again
$ x+3=4\ne 5 $
So, x=1 is not our solution again.
We put x=2
We get –
$ \begin{align}
& x+3=2+3 \\
& =5 \\
\end{align} $
So, $ x+3=5 $
So, x=2 is our required solution.
This method is a bit extensive.
Another way is to go by algebra.
We have $ x+3=5 $
We subtract -3 from both sides, we get –
So, we get –
$ x+3-3=5-3 $
As $ 3-3=0 $ and $ 5-3=2 $
So, we get –
$ x=2 $
So, solution is $ x=2 $
We get –
For $ x+3=5 $
Solution is $ x=2 $
Note:
While solving this problem hit and the trial will become more lengthy as sometimes we may start from a point and more along say positive director but our solution lie into the 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.
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