Question

Solve for x. 1/a+b+x = 1/a + 1/b + 1/x (where a+b+c is not =0 and a,b,x is not = 0) if u dont understand or else see the image there is the question plzzz answer it to me guysss
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Answer 1

Answer

x = -a, - b

$\rule{200}2$

Explanation-

$\implies\:\sf{\dfrac{1}{a+b+x}\:=\:\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{x}}$

$\implies\:\sf{\frac{1}{a+b+x}\:=\:\frac{1}{a}+\frac{1}{b}+\frac{1}{x}}$

$\implies\:\sf{\frac{1}{a+b+x}-\frac{1}{x}\:=\:\frac{1}{a}+\frac{1}{b}}$

$\implies\:\sf{\frac{x-(a+b+x)}{x(a+b+x)}\:=\:\frac{1}{a}+\frac{1}{b}}$

$\implies\:\sf{\frac{-(a+b)}{x(a+b+x)}\:=\:\frac{a+b}{ab}}$

(a + b) cancel throughout

$\implies\:\sf{\frac{-1}{x(a+b+x)}\:=\:\frac{1}{ab}}$

Cross-multiply them

⇒ -ab = x² + (a + b)x

⇒ x² + (a + b)x + ab = 0

The above equation is in the form, ax² + bx + c = 0. So, we can solve them by Quadratic equation or by splitting the middle term.

Where, a = 1, b = (a + b) and c = ab

Using Quadratic formula

D = √(b² - 4ac)

⇒ D = √[(a + b)² - 4(1)(ab)]

⇒ D = √(a² + b² + 2ab - 4ab)

⇒ D = √(a² + b² - 2ab)

⇒ D = √(a - b)²

⇒ D = (a - b)

Now, x = $\sf{\frac{-b \pm D}{2a}}$

$\implies\:\sf{\frac{ {-(a+b)} \pm (a-b) }{2(1)}}$

$\implies\:\sf{\frac{ {-a-b} + (a-b) }{2}}$

$\implies\:\sf{\frac{-2b}{2}\:=\:-b}$

Similarly,

$\implies\:\sf{\frac{ {-a-b} - (a-b) }{2}}$

$\implies\:\sf{\frac{ -a-b - a+b }{2}}$

$\implies\:\sf{\frac{-2a}{2}\:=\:-a}$

Therefore,

$\implies\:\sf{x\:=\:-a, \:-b}$

Answer 2

❥${\underline{\underline{\huge{\mathtt{Question:-}}}}}$

$if \: \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}$

(where a+b+x≠0 and a,b,x≠ 0)

Then find the value of X.

❥${\underline{\underline{\huge{\mathtt{Solution:-}}}}}$

$\frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \\ = > \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \\ = > \frac{x - (a + b + x)}{(a + b + x)x} = \frac{b + a}{ab} \\ = > \frac{x - a - b - x}{xa + bx + {x}^{2} } = \frac{b + a}{ab} \\ = > \frac{ - (a + b)}{xa + bx + {x}^{2} } = \frac{b + a}{ab} \\ = > \frac{ - 1}{xa + xb + {x}^{2} } = \frac{1}{ab} \\ = > xa + xb + {x}^{2} = - ab \\ = > xa + xb + {x}^{2} + ab = 0 \\ = > {x}^{2} + xa + xb + ab = 0 \\ = > x(x + a) + b(x + a) = 0 \\ = > (x + a)(x + b) = 0$

Either,

x+a = 0

→ x = -a

Or,

x+b = 0

→ x = -b

❥${\underline{\underline{\huge{\mathtt{Answer:-}}}}}$

X = -a, -b