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Solve each of the following equations and also verify your solution:
$\dfrac{x}{2}+\dfrac{x}{8}=\dfrac{1}{8}$.

Answer
VerifiedVerified
482.7k+ views
Hint: In this question, we have been given an equation, which we can solve by adding the fractions on the left hand side and then we will cross multiply these two fractions to get the value of the variable. Then to verify our solution, we will put the obtained value of the variable in the left hand side of the equation.

Complete step-by-step answer:
The given equation is
$\frac{x}{2}+\dfrac{x}{8}=\dfrac{1}{8}$
Now, we will add both the fractions on the left hand side of the equation.
$\Rightarrow \dfrac{4x+x}{8}=\frac{1}{8}$
On adding the terms of the numerator, we get
$\Rightarrow \dfrac{5x}{8}=\dfrac{1}{8}$
On cross multiplying the terms, we get
$\Rightarrow 8\times 5x=8$
Dividing both sides by 8, we get
$\begin{align}
  & \Rightarrow \dfrac{8\times 5x}{8}=\dfrac{8}{8} \\
 & \Rightarrow 5x=1 \\
\end{align}$
Now, we will divide 5 on both sides.
$\Rightarrow \dfrac{5x}{5}=\dfrac{1}{5}$
On further simplification, we get
$\Rightarrow x=\dfrac{1}{5}$
Thus, the value of $x$ is $\dfrac{1}{5}$.
Now, we will verify our solution. To verify the solution, we will put the obtained value of $x$ in the given equation.
$\Rightarrow \dfrac{\dfrac{1}{5}}{2}+\dfrac{\dfrac{1}{5}}{8}=\dfrac{1}{8}$
On further simplification, we get
$\Rightarrow \dfrac{1}{10}+\dfrac{1}{40}=\dfrac{1}{8}$
On adding the two fractions on left side of equation, we get
$\Rightarrow \dfrac{4+1}{40}=\dfrac{1}{8}$
On adding the terms of the numerator, we get
$\Rightarrow \dfrac{5}{40}=\dfrac{1}{8}$
On simplification, we get
$\Rightarrow \dfrac{1}{8}=\dfrac{1}{8}$
We can see that the value of LHS and RHS of the given are the same. Hence, verified.

Note: Here we have added two fractions whose denominators are different. So to add or subtract such fractions, we first make the denominator of the fractions the same by multiplying the required number to both numerator and denominator and then we simply add the terms of the numerator by keeping the denominator the same. If any non-zero term is multiplied to a variable in the equation, then we simply divide both sides by that number to get the value of the variable.
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