If \[98x=2\]. Then find the value of x.
Answer
382.2k+ views
Hint: Take LHS = 98x and RHS = 2. Take 2 to LHS, it becomes an algebraic expression. Simplify the expression received and solve the entity to obtain the value of x.
Complete step-by-step answer:
An algebraic expression is an expression built up from integers, constants, variables and exponentiation by an exponentiation by an exponent that is a rational number.
Given is the expression,
\[98x=2\]
Taking 2 to the LHS, we get
\[98x-2=0\]
Taking 2 common on LHS,
\[2\left( 49x-1 \right)=0\]
\[\therefore \] We get \[49x-1=0\]
\[\begin{align}
& \therefore 49x=1 \\
& x=\dfrac{1}{49} \\
\end{align}\]
Hence, we got the value of x as \[\dfrac{1}{49}\].
Note:
The expression can be solved directly.
Take 98 to the denominator of RHS. 98 is a multiple of 2. So 98 has a common factor. So 98 can be written as \[2\times 49\], which is equal to 98.
\[\begin{align}
& \therefore 98x=2 \\
& \Rightarrow x=\dfrac{2}{98}=\dfrac{2}{2\times 49} \\
\end{align}\]
Cancel out 2 on the numerator and denominator.
\[\therefore x=\dfrac{1}{49}\].
Complete step-by-step answer:
An algebraic expression is an expression built up from integers, constants, variables and exponentiation by an exponentiation by an exponent that is a rational number.
Given is the expression,
\[98x=2\]
Taking 2 to the LHS, we get
\[98x-2=0\]
Taking 2 common on LHS,
\[2\left( 49x-1 \right)=0\]
\[\therefore \] We get \[49x-1=0\]
\[\begin{align}
& \therefore 49x=1 \\
& x=\dfrac{1}{49} \\
\end{align}\]
Hence, we got the value of x as \[\dfrac{1}{49}\].
Note:
The expression can be solved directly.
Take 98 to the denominator of RHS. 98 is a multiple of 2. So 98 has a common factor. So 98 can be written as \[2\times 49\], which is equal to 98.
\[\begin{align}
& \therefore 98x=2 \\
& \Rightarrow x=\dfrac{2}{98}=\dfrac{2}{2\times 49} \\
\end{align}\]
Cancel out 2 on the numerator and denominator.
\[\therefore x=\dfrac{1}{49}\].
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