Question

Solve each of the following systems of equations by the method of cross-multiplication:
$bx + cy = a + c$
$ax\left( {\dfrac{1}{{a - b}} - \dfrac{1}{{a + b}}} \right) + cy\left( {\dfrac{1}{{b - a}} - \dfrac{1}{{b + a}}} \right) = \dfrac{{2a}}{{a + b}}$

Answer 1

Hint: In order to solve this problem we need to use the formula of cross multiplication also known as cramer's rule the formula required here for the two of the equation \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] so, we do $\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$

Complete step-by-step answer:
Given equations are:
$bx + cy = a + c$
$ax\left( {\dfrac{1}{{a - b}} - \dfrac{1}{{a + b}}} \right) + cy\left( {\dfrac{1}{{b - a}} - \dfrac{1}{{b + a}}} \right) = \dfrac{{2a}}{{a + b}}$
The above equations can be written as,
$bx + cy - a - c = 0$ …………………(1)
$ax\left( {\dfrac{1}{{a - b}} - \dfrac{1}{{a + b}}} \right) + cy\left( {\dfrac{1}{{b - a}} - \dfrac{1}{{b + a}}} \right) - \dfrac{{2a}}{{a + b}} = 0$………………..(2)
On simplifying the second equation,
$
  ax\left( {\dfrac{{a + b - a + b}}{{{a^2} - {b^2}}}} \right) + cy\left( {\dfrac{{b + a - b + a}}{{{b^2} - {a^2}}}} \right) = \dfrac{{2a}}{{a + b}} \\
  ax\left( {\dfrac{{2b}}{{{a^2} - {b^2}}}} \right) + cy\left( {\dfrac{{2a}}{{{b^2} - {a^2}}}} \right) = \dfrac{{2a}}{{a + b}} \\
  ax\left( {\dfrac{{2b}}{{{a^2} - {b^2}}}} \right) - cy\left( {\dfrac{{2a}}{{{a^2} - {b^2}}}} \right) = \dfrac{{2a}}{{a + b}} \\
  bx - cy = a - b \\
$
Hence, the second equation can be written as
$bx - cy = a - b$…………..(3)
We know that if there are two equations like \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] then on solving it by cross multiplications method we need to use the formula $\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$…………(1)
Here,
$
  {a_1} = b,\,{b_1} = c,\,{c_1} = a + c \\
  {a_2} = b,\,{b_2} = - c,\,{c_2} = a - c \\
$
So, on putting these values in equation (1) we get,
$
   \Rightarrow \dfrac{x}{{\left( c \right)\left( {a - c} \right) - \left( { - c} \right)\left( {a + c} \right)}} = \dfrac{y}{{\left( {a + c} \right)\left( b \right) - \left( {a - c} \right)\left( b \right)}} = \dfrac{1}{{\left( b \right)\left( { - c} \right) - \left( b \right)\left( c \right)}} \\
   \Rightarrow \dfrac{x}{{ac - {c^2} + ac - {c^2}}} = \dfrac{y}{{ab + cb - ab + bc}} = \dfrac{1}{{ - bc - bc}} \\
   \Rightarrow \dfrac{x}{{ac}} = \dfrac{y}{{bc}} = \dfrac{1}{{ - 2bc}} \\
$
Now on equating terms one by one we will get the value of x, y.
$
   \Rightarrow \dfrac{x}{{ac}} = - \dfrac{1}{{2bc}} \\
   \Rightarrow x = - \dfrac{a}{{2b}} \\
$
So, the value of x is $ - \dfrac{a}{{2b}}$.
Now on finding the value of y we get,
$
   \Rightarrow \dfrac{y}{{bc}} = - \dfrac{1}{{2bc}} \\
   \Rightarrow y = - \dfrac{1}{2} \\
$
Hence, the value of y is $ - \dfrac{1}{2}$
So, the equations $bx + cy = a + c$,
$ax\left( {\dfrac{1}{{a - b}} - \dfrac{1}{{a + b}}} \right) + cy\left( {\dfrac{1}{{b - a}} - \dfrac{1}{{b + a}}} \right) = \dfrac{{2a}}{{a + b}}$ has been solved by finding the value of x and y.
The value of x is $ - \dfrac{a}{{2b}}$ and that of y is $ - \dfrac{1}{2}$.

Note: Whenever you get to solve such problems you need to know that if it is said to solve the equations present in the question then you have to find the value of the variable present in the equations. Here it is asked to solve the equations by the method of cross multiplication so we have used Cramer's rule to find the values of variables. If the equations are \[{a_1}x + {b_1}y + {c_1} = 0\] and \[{a_2}x + {b_2}y + {c_2} = 0\] so, we do $\dfrac{x}{{{b_1}{c_2} - {b_2}{c_1}}} = \dfrac{y}{{{c_1}{a_2} - {c_2}{a_1}}} = \dfrac{1}{{{a_1}{b_2} - {a_2}{b_1}}}$. Here students need to take care of the sequence written in the formula changing that will give you the wrong answer. Doing this will solve your problem.