Question

How do you solve $\dfrac{7}{{a - 4}} = \dfrac{{14}}{6}$?

Answer 1

Hint: In order to determine the value of variable $a$ in the above equation, first multiply both sides of the equation with $\left( {a - 4} \right)\left( 6 \right)$ and cancel out all the possible terms. Simplify further by transposing constant terms towards one side of the equation and divide both sides of the equation with the coefficient of the variable $a$ to get the required solution.

Complete step-by-step solution:
We are given a linear equation in one variable $\dfrac{7}{{a - 4}} = \dfrac{{14}}{6}$.and we have to solve this equation for variable ($a$).
To simplify the equation, multiply both sides of the equation with the $\left( {a - 4} \right)\left( 6 \right)$, we get
$\Rightarrow \left( {a - 4} \right)\left( 6 \right) \times \dfrac{7}{{a - 4}} = \left( {a - 4} \right)\left( 6 \right) \times \dfrac{{14}}{6}$
Now cancelling out all the possible terms from both of the sides of the equation, we get
$\Rightarrow \left( 6 \right) \times 7 = \left( {a - 4} \right)14$
Now applying the distributive property of multiplication on the right-hand side of the equation to expand the term as $A\left( {B + C} \right) = AB + AC$and multiplying the values on the left-hand side , we get our equation as
$\Rightarrow 42 = 14a - 56$
Now transposing all the constant terms from the RHS to LHS using the rules of transposing of terms.
Remember that when a term is transposed from any side towards another side the sign of the term gets reversed. So, When $ - 56$is transposed from RHS to LHS it will become $56$.
$\Rightarrow 42 + 56 = 14a$
Simplifying the above further we get
$\Rightarrow 98 = 14a$
Now dividing both sides of the equation with the coefficient of the variable $a$i.e.$14$
$
  \Rightarrow \dfrac{{14a}}{{14}} = \dfrac{{98}}{{14}} \\
  \Rightarrow a = 7 \\
 $

Therefore, we have obtained the solution of the given equation as $a = 7$.

Additional Information:
i) Linear Equation in one variable: A linear equation is an equation which can be represented in the form of $ax + c$ where $x$ is the unknown variable and a,c are the numbers known where $a \ne 0$. If $a = 0$ then the equation will become constant value and will no more be a linear equation.
ii) The degree of the variable in the linear equation is of the order 1.
iii) Every Linear equation has 1 root.

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.There is only one value of $a$ which is the solution to the equation and if we put this $a$ in the equation, the equation will be zero.
3.Like terms are the terms who have the same variable and power.
4. Graph of the linear equation is always a straight line.