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Find the value of $x$:
$ 9x+71=42 x+98 $
Given:
\( 9x+71=42 x+98 \)
To do:
We have to find the value of $x$.
Solution:
\( 9x+71=42 x+98 \)
$\Rightarrow 42x-9x=71-98$
$\Rightarrow 33x=-27$
$\Rightarrow x=\frac{-27}{33}$
$\Rightarrow x=\frac{-9}{11}$
The value of $x$ is $\frac{-9}{11}$.
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