Question

Slove for X \frac{1}{a} + \frac{1}{b} + \frac{1}{x} = \frac{1}{a+b+x}

Answer 1

Answer:

- a, -b

Step-by-step explanation:

Given:

$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{x} = \dfrac{1}{a+b+x}$

To find:

The value of x

Solution:

It's seem to be a equation but it's a quadratic equation.

We will follow the some rule.

i) sign of any no. or variable will change between equal (=) like + will change in - and multiple will change in division.

ii) We can simply eliminate the no. from one side to another side if they are in multiple or have common factor.

Now,

$\frac{1}{a} + \frac{1}{b} + \frac{1}{x} = \frac{1}{a+b+x} \\ \\ \frac{1}{a} + \frac{1}{b} = \frac{1}{a+b+x} - \frac{1}{x} \\ \\ \frac{b + a}{ab} = \frac{x - (a + b + x)}{x(a + b + x)} \: \: [\text{By Taking L.C.M}] \\ \\ \frac{a + b}{ab} = \frac{ \cancel{x} - a - b - \cancel{x}}{ax + bx + x {}^{2} } \\ \\ \frac{ \cancel{(a + b)}}{ab} = \frac{ - \cancel{(a + b)}}{ax + bx + x {}^{2} } \\ \\ ax + bx + x {}^{2} = - ab\: \: [\text{By Cross Multiply}] \\ \\ {x}^{2} + ax + bx + ab = 0 \\ \\ x(x + a) + b(x + a) = 0 \\ \\ (x + a)(x + b) = 0 \\ \\ x + a = 0 \\ \\ x = - a \\ \\ x + b = 0 \\ \\ x = - b$

Therefore, the required value of x is -a and -b.