Answer
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Hint: Take all the variable terms of x in LHS and constants to the RHS of the equation. In this equation, just take the 2x term to LHS and 3 to RHS and solve it further.
Complete step-by-step answer:
According to the question, we have a linear equation. In that linear equation, we have a variable x which is unknown. We have to find the value of x using the given equation.
We have,
\[8x-3=9-2x\]…………(1)
In LHS we have unknown variable x terms and constant terms.
Similarly, we have variable x terms and constant terms in RHS of the equation.
So, our target is to make one side of the equation to have only unknown variable x terms.
In equation(1), taking the term -2x to the LHS, we get
\[8x-3+2x=9\]……………….(2)
Now, in the equation (2), taking -3 to the RHS, we get
\[\begin{align}
& 8x+2x=9+3 \\
& \Rightarrow 10x=12 \\
& \Rightarrow x=\dfrac{12}{10}=\dfrac{6}{5} \\
\end{align}\]
Hence, the value of x is \[\dfrac{6}{5}\] .
Note: In this question, one can make mistakes in taking -2x of RHS to the LHS and write -2x in LHS too, which is wrong. If we are moving some terms of RHS to LHS then its sign is changed and vice-versa. Here we are moving -2x to the LHS. So, we have to write +2x in the LHS of the equation. Similarly, we are moving -3 of LHS to the RHS. So, we have to write +3 in the RHS of the equation.
Complete step-by-step answer:
According to the question, we have a linear equation. In that linear equation, we have a variable x which is unknown. We have to find the value of x using the given equation.
We have,
\[8x-3=9-2x\]…………(1)
In LHS we have unknown variable x terms and constant terms.
Similarly, we have variable x terms and constant terms in RHS of the equation.
So, our target is to make one side of the equation to have only unknown variable x terms.
In equation(1), taking the term -2x to the LHS, we get
\[8x-3+2x=9\]……………….(2)
Now, in the equation (2), taking -3 to the RHS, we get
\[\begin{align}
& 8x+2x=9+3 \\
& \Rightarrow 10x=12 \\
& \Rightarrow x=\dfrac{12}{10}=\dfrac{6}{5} \\
\end{align}\]
Hence, the value of x is \[\dfrac{6}{5}\] .
Note: In this question, one can make mistakes in taking -2x of RHS to the LHS and write -2x in LHS too, which is wrong. If we are moving some terms of RHS to LHS then its sign is changed and vice-versa. Here we are moving -2x to the LHS. So, we have to write +2x in the LHS of the equation. Similarly, we are moving -3 of LHS to the RHS. So, we have to write +3 in the RHS of the equation.
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