Solve the following equation and check your result:
\[5x+9=5+3x\]
Answer
609.9k+ views
Hint: We solve this problem by using the interchanging of terms.
We change the terms from LHS to RHS or RHS to LHS so that we get all the variables on one side and the constants on the other side which will be easy to calculate the variable.
Then we check the solution we got by substituting the solution in the given equation.
Complete step by step answer:
We are given that the equation as
\[5x+9=5+3x\]
Now, let us use the interchanging of terms method to solve this problem.
Let us change the terms in such a way that all variables get one side and the constants get one side.
We know that during the process of changing the terms the addition becomes subtraction.
By using the above theorem to given equation we get
\[\begin{align}
& \Rightarrow 5x-3x=5-9 \\
& \Rightarrow 2x=-4 \\
& \Rightarrow x=-2 \\
\end{align}\]
Therefore we can conclude that the solution of given equation is \[x=-2\]
Now, let us check whether the solution we got is correct or not.
We are given that the equation as
\[\Rightarrow 5x+9=5+3x\]
Let us take the LHS as
\[\Rightarrow LHS=5x+9\]
By substituting \[x=-2\] in above equation we get
\[\begin{align}
& \Rightarrow LHS=5\left( -2 \right)+9 \\
& \Rightarrow LHS=-10+9 \\
& \Rightarrow LHS=-1.....equation(i) \\
\end{align}\]
Now, let us take the RHS of given equation as
\[\Rightarrow RHS=5+3x\]
By substituting \[x=-2\] in above equation we get
\[\begin{align}
& \Rightarrow RHS=5+3\left( -2 \right) \\
& \Rightarrow RHS=5-6 \\
& \Rightarrow RHS=-1.....equation(ii) \\
\end{align}\]
Now, by comparing the equation (i) and equation (ii) we get
\[\Rightarrow LHS=RHS\]
Note: Students may make mistakes sometimes without checking if the solution we got is correct or wrong.
Here we got the solution of given equation as
\[\Rightarrow x=-2\]
After this we need to substitute this solution in the given equation to check whether the solution is correct or wrong.
Sometimes we may get the solution in the process where it doesn’t satisfy the given equation.
We need to check every time after solving the variable.
We change the terms from LHS to RHS or RHS to LHS so that we get all the variables on one side and the constants on the other side which will be easy to calculate the variable.
Then we check the solution we got by substituting the solution in the given equation.
Complete step by step answer:
We are given that the equation as
\[5x+9=5+3x\]
Now, let us use the interchanging of terms method to solve this problem.
Let us change the terms in such a way that all variables get one side and the constants get one side.
We know that during the process of changing the terms the addition becomes subtraction.
By using the above theorem to given equation we get
\[\begin{align}
& \Rightarrow 5x-3x=5-9 \\
& \Rightarrow 2x=-4 \\
& \Rightarrow x=-2 \\
\end{align}\]
Therefore we can conclude that the solution of given equation is \[x=-2\]
Now, let us check whether the solution we got is correct or not.
We are given that the equation as
\[\Rightarrow 5x+9=5+3x\]
Let us take the LHS as
\[\Rightarrow LHS=5x+9\]
By substituting \[x=-2\] in above equation we get
\[\begin{align}
& \Rightarrow LHS=5\left( -2 \right)+9 \\
& \Rightarrow LHS=-10+9 \\
& \Rightarrow LHS=-1.....equation(i) \\
\end{align}\]
Now, let us take the RHS of given equation as
\[\Rightarrow RHS=5+3x\]
By substituting \[x=-2\] in above equation we get
\[\begin{align}
& \Rightarrow RHS=5+3\left( -2 \right) \\
& \Rightarrow RHS=5-6 \\
& \Rightarrow RHS=-1.....equation(ii) \\
\end{align}\]
Now, by comparing the equation (i) and equation (ii) we get
\[\Rightarrow LHS=RHS\]
Note: Students may make mistakes sometimes without checking if the solution we got is correct or wrong.
Here we got the solution of given equation as
\[\Rightarrow x=-2\]
After this we need to substitute this solution in the given equation to check whether the solution is correct or wrong.
Sometimes we may get the solution in the process where it doesn’t satisfy the given equation.
We need to check every time after solving the variable.
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