
Find the value of $x$ from the expression: $\dfrac{{x - 5}}{2} - \dfrac{{x - 3}}{5} = \dfrac{1}{2}$.
Answer
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Hint:In order to solve this question, to solve for the value of $x$ , as we can see that the given equation is a linear type equation. So, we will try to make the simplest form first on both sides as well and then we will cross multiply to solve for $x$ .
Complete step by step answer:
The given equation is:
$\dfrac{{x - 5}}{2} - \dfrac{{x - 3}}{5} = \dfrac{1}{2}$
First, we will solve for the L.H.S, by doing the LCM of the denominator:
$ \Rightarrow \dfrac{{5(x - 5) - 2(x - 3)}}{{10}} = \dfrac{1}{2}$
And now we will solve the numerator of L.H.S:-
$\dfrac{{5x - 25 - 2x + 6}}{{10}} = \dfrac{1}{2} \\
\Rightarrow \dfrac{{3x - 19}}{{10}} = \dfrac{1}{2} \\ $
Now, we can do cross-multiplication to solve for $x$ :
$2(3x - 19) = 10 \\
\Rightarrow 6x - 38 = 10 \\
\Rightarrow 6x = 10 + 38 \\
\therefore x = \dfrac{{48}}{6} = 8 \\ $
Hence, the value of $x$ for the given expression is 8.
Note: Fractions, rather than whole numbers, can be used to express linear equations. These are known as fractional linear equations. To solve equations containing fractions, we must first turn them into equations that do not contain fractions.
Complete step by step answer:
The given equation is:
$\dfrac{{x - 5}}{2} - \dfrac{{x - 3}}{5} = \dfrac{1}{2}$
First, we will solve for the L.H.S, by doing the LCM of the denominator:
$ \Rightarrow \dfrac{{5(x - 5) - 2(x - 3)}}{{10}} = \dfrac{1}{2}$
And now we will solve the numerator of L.H.S:-
$\dfrac{{5x - 25 - 2x + 6}}{{10}} = \dfrac{1}{2} \\
\Rightarrow \dfrac{{3x - 19}}{{10}} = \dfrac{1}{2} \\ $
Now, we can do cross-multiplication to solve for $x$ :
$2(3x - 19) = 10 \\
\Rightarrow 6x - 38 = 10 \\
\Rightarrow 6x = 10 + 38 \\
\therefore x = \dfrac{{48}}{6} = 8 \\ $
Hence, the value of $x$ for the given expression is 8.
Note: Fractions, rather than whole numbers, can be used to express linear equations. These are known as fractional linear equations. To solve equations containing fractions, we must first turn them into equations that do not contain fractions.
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