Question

Solve the given equation
$ \dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3 $

Answer 1

Hint: First write the equation in terms of x, a, b, c. Now take the least common multiple and solve it till you get a single fraction. Try to manipulate such that the left hand side must be with all possible x terms. Then find the value of coefficient of x. Divide with it on both sides, by this you get a value of x. This value of x is the required result.

Complete step-by-step answer:
Given equation in terms of x, a, b, c can be written as
 $ \dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3 $
By taking least common multiple in first two terms, we get
 $ \dfrac{b\left( x-b-c \right)+a\left( x-c-a \right)}{ab}+\dfrac{x-a-b}{c}=3 $
By multiplying b inside we can remove the bracket, we get it as
 $ \dfrac{bx-{{b}^{2}}-bc+a\left( x-c-a \right)}{ab}+\dfrac{x-a-b}{c}=3 $
By multiplying a inside we can remove other bracket, we get
 $ \dfrac{bx-{{b}^{2}}-bc+ax-ac-{{a}^{2}}}{ab}+\dfrac{x-a-b}{c}=3 $
Again by taking least common multiple we get it as
 $ \dfrac{c\left( bx-{{b}^{2}}-bc+ax-ac-{{a}^{2}} \right)+ab\left( x-a-b \right)}{abc}=3 $
By cross multiplying we can write the above equation as
 $ c\left( bx-{{b}^{2}}-bc+ax-ac-{{a}^{2}} \right)+ab\left( x-a-b \right)=3abc $
By multiplying c inside we can remove bracket and write it as
 $ bcx-{{b}^{2}}-b{{c}^{2}}+acx-a{{c}^{2}}-{{a}^{2}}c+ab\left( x-a-b \right)=3abc $
By multiplying the term ab inside, we can write it in form of
 $ bcx-{{b}^{2}}c-b{{c}^{2}}+acx-{{a}^{2}}c-a{{c}^{2}}+abx-{{a}^{2}}b-a{{b}^{2}}=3abc $
By adding $ \left( {{b}^{2}}c+b{{c}^{2}}+{{a}^{2}}c+a{{c}^{2}}+{{a}^{2}}b+a{{b}^{2}} \right) $ we get in form of
 $ \begin{align}
  & \left( bcx+acx+abx \right)-\left( {{b}^{2}}c+b{{c}^{2}}+{{a}^{2}}c+a{{c}^{2}}+{{a}^{2}}b+a{{b}^{2}} \right)+\left( {{b}^{2}}c+b{{c}^{2}}+{{a}^{2}}c+a{{c}^{2}}+{{a}^{2}}b+a{{b}^{2}} \right) \\
 & =3abc+\left( {{b}^{2}}c+b{{c}^{2}}+{{a}^{2}}c+a{{c}^{2}}+{{a}^{2}}b+a{{b}^{2}} \right) \\
\end{align} $
By cancelling the common terms, we can write the equation in form of
 $ bcx+acx+abx=3abc+{{b}^{2}}c+b{{c}^{2}}+{{a}^{2}}c+a{{c}^{2}}+{{a}^{2}}b+a{{b}^{2}} $
By taking x from left hand side, 3abc writing it as abc 3times, we get
 $ \left( ab+bc+ac \right)x=b{{c}^{2}}+{{b}^{2}}c+abc+{{a}^{2}}c+a{{c}^{2}}+abc+{{a}^{2}}b+a{{b}^{2}}+abc $
Taking bc common from first three terms, we get it in form of
 $ \left( ab+bc+ac \right)x=bc\left( a+b+c \right)+{{a}^{2}}c+a{{c}^{2}}+abc+{{a}^{2}}b+a{{b}^{2}}+abc $
Similarly take ac, ab from next three terms, we get it as
 $ \left( ab+bc+ac \right)x=bc\left( a+b+c \right)+ac\left( a+b+c \right)+ab\left( a+b+c \right) $
By taking a + b + c common on right hand side, we get it as
 $ \left( ab+bc+ac \right)x=\left( ab+bc+ca \right)\left( a+b+c \right) $
By cancelling common terms, we get the value of x to be
x = a + b + c.
Therefore this is the solution of the equation in question.

Note: Be careful while multiplying terms inside the bracket. Generally students forget to multiply with last terms, which may lead to wrong answers. The idea of bringing all x to the left hand side is because we want to take x as common and calculate its value. We take bc, ab, ca as common to generate the term ab + bc + ca which is present in the left hand side of the equation and use the 3abc term properly.