Solve the following:
(a) $2\left( 5x-3 \right)-3\left( 2x-1 \right)=9$
(b) $\dfrac{x}{2}=\dfrac{x}{3}+1$
(c) $\dfrac{x}{2}+\dfrac{x}{3}=\dfrac{2x}{5}-1$
Answer
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Hint: For answering this question we will use the given expression in each part and arrange all the terms containing $x$ to the left hand side from the right hand side and then simplify it as required per situation and get the value of $x$ .
Complete step by step answer:
Now considering from the question we have the given expression in the first part is $2\left( 5x-3 \right)-3\left( 2x-1 \right)=9$ .
By simplifying the left hand side we will have $10x-6-6x+3=9$ .
By further simplifying we will have $4x-3=9$.
After transferring the like terms aside, that is the non $x$ terms on the right hand side we will have $4x=9+3$ .
So we will have $4x=12$ .
Hence we will have $x=3$ for the first part.
Now considering from the question we have the given expression in the second part is $\dfrac{x}{2}=\dfrac{x}{3}+1$ .
After transferring the like terms aside, that is the $x$ terms on the left hand side we will have $\dfrac{x}{2}-\dfrac{x}{3}=1$ .
By further simplifying the left hand side we will have $\dfrac{3x-2x}{6}=1$ .
By simplifying this we will have $x=6$.
Hence we will have $x=6$ for the second part.
Now considering from the question we have the given expression in the third part is $\dfrac{x}{2}+\dfrac{x}{3}=\dfrac{2x}{5}-1$ .
After transferring the like terms aside that is the $x$ terms on left hand side we will have $\dfrac{x}{2}+\dfrac{x}{3}-\dfrac{2x}{5}=1$ .
By further simplifying the left hand side we will have $\dfrac{3x+2x}{6}-\dfrac{2x}{5}=1$ .
By simplifying this we will have
$\begin{align}
& \dfrac{5x}{6}-\dfrac{2x}{5}=1 \\
& \Rightarrow \dfrac{25x-12x}{30}=1 \\
\end{align}$ .
So we will have $\dfrac{13x}{30}=1$ .
Hence we will have $x=\dfrac{30}{13}$ for the third part.
Note: While answering questions of this type we should take care while performing the calculations for example in the third part if we had made a mistake and written as $\dfrac{25x-12x}{30}=1\Rightarrow \dfrac{12x}{30}=1\Rightarrow x=\dfrac{5}{2}$ which is clearly wrong. The important point that we use here is that the left hand side and the right hand side of an expression are always equal to each other. We can substitute the value of x and then check if the right hand side and left hand side are the same or not.
Complete step by step answer:
Now considering from the question we have the given expression in the first part is $2\left( 5x-3 \right)-3\left( 2x-1 \right)=9$ .
By simplifying the left hand side we will have $10x-6-6x+3=9$ .
By further simplifying we will have $4x-3=9$.
After transferring the like terms aside, that is the non $x$ terms on the right hand side we will have $4x=9+3$ .
So we will have $4x=12$ .
Hence we will have $x=3$ for the first part.
Now considering from the question we have the given expression in the second part is $\dfrac{x}{2}=\dfrac{x}{3}+1$ .
After transferring the like terms aside, that is the $x$ terms on the left hand side we will have $\dfrac{x}{2}-\dfrac{x}{3}=1$ .
By further simplifying the left hand side we will have $\dfrac{3x-2x}{6}=1$ .
By simplifying this we will have $x=6$.
Hence we will have $x=6$ for the second part.
Now considering from the question we have the given expression in the third part is $\dfrac{x}{2}+\dfrac{x}{3}=\dfrac{2x}{5}-1$ .
After transferring the like terms aside that is the $x$ terms on left hand side we will have $\dfrac{x}{2}+\dfrac{x}{3}-\dfrac{2x}{5}=1$ .
By further simplifying the left hand side we will have $\dfrac{3x+2x}{6}-\dfrac{2x}{5}=1$ .
By simplifying this we will have
$\begin{align}
& \dfrac{5x}{6}-\dfrac{2x}{5}=1 \\
& \Rightarrow \dfrac{25x-12x}{30}=1 \\
\end{align}$ .
So we will have $\dfrac{13x}{30}=1$ .
Hence we will have $x=\dfrac{30}{13}$ for the third part.
Note: While answering questions of this type we should take care while performing the calculations for example in the third part if we had made a mistake and written as $\dfrac{25x-12x}{30}=1\Rightarrow \dfrac{12x}{30}=1\Rightarrow x=\dfrac{5}{2}$ which is clearly wrong. The important point that we use here is that the left hand side and the right hand side of an expression are always equal to each other. We can substitute the value of x and then check if the right hand side and left hand side are the same or not.
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