Question

if the equations $a(y + z) = x$$,b(z + x) = y,c(x + y) = x$ have non-trivial solution, then $\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = $
A) $1$
B) $2$
C) $-1$
D) $-2$

Answer 1

Hint: These have non trivial solutions for non- trivial solutions first we will understand what are non- trivial solutions. Solutions or examples involving the number zero are considered as trivial. Non- zero solutions or examples are considered non trivial solutions. For example the equation $x + 5y = 0$ has the solution $(0,0)$ while non-trivial solution include $(5, - 1)$ and $( - 2,0.4)$ hence this equation will also have non zero value of solution to find the value of such equations firstly we will have to rearrange each equation to find its value the rearrangement will be done according to the requirement of the equation because each equation can be rearranged in different ways after rearrangement. By adding $1$ to each side and doing the reciprocal of the fraction obtained we will proceed the solution. After obtaining three different equations we can find their sum to obtain our desired result

Complete step-by-step answer:
in the given equation firstly we will rearrange the equation on rearranging we will get
$a = \dfrac{x}{{y + z}}$ ; $b = \dfrac{y}{{z + x}}$ ; $c = \dfrac{z}{{x + y}}$
To form the equation in such a way that it may form like $\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = $
We will add 1 to both sides
$a + 1 = \dfrac{x}{{y + z}} + 1$
By taking L.C.M OF $y + z$
$a + 1 = \dfrac{{x + y + z}}{{y + z}}$ …..(1)
For second equation
$b + 1 = \dfrac{y}{{z + x}} + 1$
By taking LCM OF $z + x$
$b + 1 = \dfrac{{x + y + z}}{{z + x}}$ …..(2)
For third equation
 $c + 1 = \dfrac{z}{{x + y}} + 1$
Taking LCM $x + y$
$c + 1 = \dfrac{{z + x + y}}{{x + y}}$ ……(3)
In order to form the equations and find the value as it is given so on doing reciprocal of equation 1,2,3
$\dfrac{1}{{(a + 1)}} = \dfrac{{y + z}}{{x + y + z}}$ ..…(4)
Second equation will be
$\dfrac{1}{{(b + 1)}} = \dfrac{{z + x}}{{x + y + z}}$ …..(5)
Third equation will be
$\dfrac{1}{{(1 + c)}} = \dfrac{{x + y}}{{z + x + y}}$ ……(6)
On adding the equations 4,5,6
$\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = \dfrac{{y + z}}{{x + y + z}} + \dfrac{{z + x}}{{x + y + z}} + \dfrac{{x + y}}{{x + y + z}}$
Taking $(x + y + z)$ as LCM and adding
$\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = \dfrac{{y + z + z + x + x + y}}{{(x + y + z)}}$
We will get
$\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = \dfrac{{2(x + y + z)}}{{(x + y + z)}}$
$\dfrac{1}{{1 + a}} + \dfrac{1}{{1 + b}} + \dfrac{1}{{1 + c}} = 2$

So, the correct answer is “Option B”.

Note: In such type of solutions students mainly get confused how to form the equation to solve they do the wrong rearrangement sometimes in cases by doing reciprocals or by adding or subtracting any number according to the need of the equation in this case we have added $1$ to both sides to get the desired equation similarly for other equation we do such things