Question

solve for x 1/a+b+x=1/a+1/b+1/x where a+b+x= 0​

Answer 1

Answer:

x = -2b or -2a

Step-by-step explanation:

1/(a+b+x)=1/a + 1/b + 1/x

1/(a+b+x) -1/x = 1/a + 1/b

x-(a+b+x)/x(a+b+x) = a+b/ ab

-1(a+b) /x(a+b+x) = a+b/ ab

-1(a+b)/(a+b) = x(a+b+x)/ab

-1=x(a+b+x) / ab

-ab= (a+b)x + x2

so, x2 + (a+b)x +ab = 0

now, A = 1, B=(a+b) C= ab..

so, D = B2 - 4AC

 =(a+b)2 - 4ab

 = (a-b)2

so, x = (-B + root D)/2A  and (-B-root D)/2A

 = -a-b+a-b /2*1   and  -a-b-a+b /2*1

 = -2b/1  and -2a /1

so,x = -2b or -2a

Answer 2

Answer:

The answer is x= -a, -b

Step-by-step explanation:

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

Taking terms containing x on one side

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

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

$\frac{x - a - b - x }{ax + bx + {x}^{2} } = \frac{a + b}{ab}$

$\frac{ - (a + b)}{ax + bx + {x}^{2} } = \frac{a + b}{ab}$

Clearly, a+b can be cancelled out,

$\frac{1}{ax + bx + {x}^{2} } = \frac{ - 1}{ab}$

On cross multiplying,

$ab = - (ax + bx + {x}^{2} )$

$ab = - ax - bx - {x}^{2}$

Taking all terms on LHS,

${x}^{2} + ax + bx + ab = 0$

On taking out common factors,

$x(x + a) + b(x + a) = 0$

Taking (x+a) as common,

$(x + a)(x + b) = 0$

This would give us

$x = - a \: or \: x = - b$