Question

Solve the following for the value of x: $\dfrac{15}{4}-7x=9$ .

Answer 1

Hint: To solve $\dfrac{15}{4}-7x=9$ , we will have to rearrange the terms such that the constant terms will be on the RHS. Then we will have to simplify the RHS and take the coefficient of x to the RHS. On further simplification, we will get the value of x.

Complete step by step solution:
We have to solve $\dfrac{15}{4}-7x=9$ . This means, we have to find the value of x. First, let us collect the constant terms on the RHS. For this, we have to move $\dfrac{15}{4}$ to the RHS. When we move a positive number from one side to the other, it will become negative.
$\Rightarrow -7x=9-\dfrac{15}{4}$
Let us simplify the RHS by taking the LCM of the denominator of each term. We will get the LCM as 4. Hence, we can write the above equation as
$\Rightarrow -7x=\dfrac{9\times 4}{1\times 4}-\dfrac{15}{4}$
We can write the above equation as
$\begin{align}
  & \Rightarrow -7x=\dfrac{36}{4}-\dfrac{15}{4} \\
 & \Rightarrow -7x=\dfrac{36-15}{4} \\
\end{align}$
Let us subtract the numerator of the RHS.
$\Rightarrow -7x=\dfrac{21}{4}$
Now, we have to take -7 from the LHS to RHS. Since it is a product, it will be the divisor in the RHS after we move it.
$\Rightarrow x=\dfrac{21}{4\times -7}$
We can simplify the denominator of RHS.
$\Rightarrow x=\dfrac{21}{-28}$
We can write the above equation as
$\Rightarrow x=-\dfrac{21}{28}$
Now, let us cancel the common factors (7) from the numerator and denominator of the RHS.
$\Rightarrow x=-\dfrac{3}{4}$

Hence, the solution of the equation $\dfrac{15}{4}-7x=9$ is $-\dfrac{3}{4}$.

Note: Students must know the rules of algebra to solve the equation. When we move a positive (or negative) term from one side to the other, it will be negative (or positive). When we move a multiplier (or divisor) from one side to the other, it will be the divisor (or multiplier).