Question

Find the value for $x$ \[\dfrac{3}{{x - 1}} + \dfrac{1}{{x - 3}} = \dfrac{4}{{x - 2}}\].

Answer 1

Hint: Here in the given question we are asked to find the value of $x$ in the given linear equation. In the given equation both Left hand side and right hand side have expressions with a variable. We can simplify these expressions on respective sides and then use transposing or cross multiplication properties to find the value of the solution $x$.

Complete step by step answer:
The given linear equation consists of only one variable and the degree of the variable
\[\dfrac{3}{{x - 1}} + \dfrac{1}{{x - 3}} = \dfrac{4}{{x - 2}}\] ------ (1 )
Thus the given equation is known as a linear equation in one variable. Firstly we take LCM in the denominator on the left hand side of equation (1)
\[ \Rightarrow \]\[\dfrac{{3\left( {x - 3} \right) + 1\left( {x - 1} \right)}}{{\left( {x - 3} \right)\left( {x - 1} \right)}} = \dfrac{4}{{x - 2}}\] -------- (2)
On simplifying the equation (2)
\[ \Rightarrow \]\[\dfrac{{3x - 9 + x - 1}}{{x\left( {x - 1} \right) - 3\left( {x - 1} \right)}} = \dfrac{4}{{x - 2}}\]
On combining like terms
\[ \Rightarrow \]\[\dfrac{{4x - 10}}{{{x^2} - x - 3x + 3}} = \dfrac{4}{{x - 2}}\]
\[ \Rightarrow \]\[\dfrac{{4x - 10}}{{{x^2} - 4x + 3}} = \dfrac{4}{{x - 2}}\] ---------- (3)

On cross multiplying the equation (3) we get
\[ \Rightarrow \] \[\left( {4x - 10} \right)\left( {x - 2} \right) = 4\left( {{x^2} - 4x + 3} \right)\] ----------- (4)
On multiplying the terms by opening the brackets
\[ \Rightarrow \] \[4x\left( {x - 2} \right) - 10\left( {x - 2} \right) = 4{x^2} - 16x + 12\] ----------- (5)
\[ \Rightarrow \] \[4{x^2} - 8x - 10x + 20 = 4{x^2} - 16x + 12\] ---------- (6)
On combining like terms in the equation (6)
\[ \Rightarrow \]\[4{x^2} - 18x + 20 = 4{x^2} - 16x + 12\]
On rearranging the terms by sending them on to one side we get
\[ \Rightarrow \]\[4{x^2} - 4{x^2} - 18x + 16x + 20 - 12 = 0\]
\[ \Rightarrow \]\[ - 2x + 8 = 0\]
On transposing 8 to RHS
\[\Rightarrow - 2x = - 8 \\
\Rightarrow 2x = 8 \\ \]
On simplifying for x we get
\[ \Rightarrow x = \dfrac{8}{2}\]
\[ \therefore x = 4\]

Thus the solution of the given linear equation in one variable is x = 4.

Note: Remember sometimes the expressions forming equations have to be simplified before we can solve them by usual methods. Some equations may not even be linear to begin with, but they can be brought to a linear form by multiplying suitable expressions. Even in some equations variables can be transposed from one side of the equation to the other.