Question

Solve the given equation:
$\dfrac{{(m + n)x - (a - b)}}{{(m - n)x - (a + b)}} = \dfrac{{(m + n)x - (a + c)}}{{(m - n)x - (a - c)}}$

Answer 1

Hint: In this question, first we add 1 on both sides of the equation and let it be equation 1 and then again in the original equation we add -1 on the both sides and let it be equation 2. Now, we will divide the equation 1 and 2 and solve it further to get the value of x.

Complete step-by-step answer:
The given equation is :
$\dfrac{{(m + n)x - (a - b)}}{{(m - n)x - (a + b)}} = \dfrac{{(m + n)x + (a + c)}}{{(m - n)x + (a - c)}}$ (1)
First, we add 1 to both sides of the above equation.
On adding 1 on both sides, we get:
$\dfrac{{(m + n)x - (a - b)}}{{(m - n)x - (a + b)}} + 1 = \dfrac{{(m + n)x + (a + c)}}{{(m - n)x + (a - c)}} + 1$
On solving it, we have:
$\dfrac{{(m + n)x - (a - b) + (m - n)x - (a + b)}}{{(m - n)x - (a + b)}} = \dfrac{{(m + n)x + (a + c) + (m - n)x + (a - c)}}{{(m - n)x + (a - c)}}$
$ \Rightarrow \dfrac{{2mx - 2a}}{{(m - n)x - (a + b)}} = \dfrac{{2mx + 2a}}{{(m - n)x + (a - c)}}$ (2)
Now, adding -1 on the both sides of equation 1, we get:
$\dfrac{{(m + n)x - (a - b)}}{{(m - n)x - (a + b)}} - 1 = \dfrac{{(m + n)x + (a + c)}}{{(m - n)x + (a - c)}} - 1$.
On rearranging the above equation, we get:
$\dfrac{{(m + n)x - (a - b) - (m - n)x + (a + b)}}{{(m - n)x - (a + b)}} = \dfrac{{(m + n)x + (a + c) - (m - n)x - (a - c)}}{{(m - n)x + (a - c)}}$
On simplifying the LHS and RHS, we get:
$ \Rightarrow \dfrac{{2nx + 2b}}{{(m - n)x - (a + b)}} = \dfrac{{2nx + 2c}}{{(m - n)x + (a - c)}}$ (3)
Now, on dividing equation 1 and 2, we get:
$\dfrac{{2mx - 2a}}{{2nx + 2b}} = \dfrac{{2mx + 2a}}{{2nx + 2c}}$
On dividing both sides by 2, we have:
$ \Rightarrow \dfrac{{mx - a}}{{nx + b}} = \dfrac{{mx + a}}{{nx + c}}$
On cross multiplying, we get:
$(mx - a)(nx + c)= (nx + b)(mx + a)$
On further solving, we get:
$mn{x^2}$ +$x(mc – an) –ac$ =$mn{x^2}$ $+x(an + bm) +ab$
Cancelling the square terms on both sides and rearranging the terms, we get:
$ \Rightarrow $ $x(mc – an) – x(an + bm) = ab +ac$.
Taking the ‘x’ common, we get:
$x(m(c –b) – 2an ) = a(b + c)$
$ \Rightarrow x = \dfrac{{a(b + c)}}{{m(c - b) - 2an}}$ .
Therefore, the above value of ‘x’ is the solution of the given equation.

Note: In such a type of question, we will use the concept of componendo and dividend. If a/b = c/d is a proportion then adding 1 on both sides is componendo and adding -1 on both sides is dividendo. To solve this question, we can also use the result $\dfrac{x}{y} = \dfrac{{x + y}}{{x - y}}$ .