Question

please answer fast ............

Answer 1

Answer:

Step-by-step explanation:

U should take -1/2a to rhs

So after taking lcm u get..

2a-2a-b-2x/[2a*(2a+b+2x)] = 2x+b/2xb

So cancel ou the needed.. then

-b-2x/[2a*(2a+b+2x)] = 2x+b/2xb

Now -(b+2x) from rhs and (2x+b) from lhs gets cancelled..

So..

-1/[2a*(2a+b+2x)] = 1/2xb

-2xb=2a*(2a+b+2x)

2 gets cancelled

-xb=2a square+ab+2ax

Now from here ull get x

...-xb-2ax=2asquare+ab

X=2asquare+ab/(-b-2a)

Answer 2

Answer: 1] condition not possible

OR part] x = -b/2 or -a

Step-by-step explanation:

1] Sum of ages = 20

Let ages 4 years ago be X , Y

Product === X*Y = 48... (i)

4 years ago sum will be 20-(4+4) = 12

X+Y = 12

X = 12-Y....(ii)

Substitute the value of X in eq. (i)

(12-Y)*Y = 48

12Y - Y^2 = 48

Y^2-12Y+48 = 0

Sum = -12 && Product = 48

There are no such 'natural numbers' that satisfy this condition and hence  the given condition is not possible

              OR

Solve for x:

1/(2a+b+2x) = 1/2a + 1/b + 1/2x

1/(2a+b+2x) - 1/2x = 1/2a + 1/b  

Taking LCM in LHS

(2x - 2a - b - 2x)/(4ax + 2bx + 4x^2) = 1/2a + 1/b

Now take LCM in RHS

(-2a-b)/(4ax + 2bx + 4x^2) = b+2a/2ab

-(2a+b)/(4ax + 2bx + 4x^2) = (2a+b)/2ab

The numerator of LHS & RHS are same, hence they cancel each other

-1/(4ax + 2bx + 4x^2) = 1/2ab

2ab = -4ax - 2bx - 4x^2

4x^2 + 2bx + 4ax + 2ab = 0

Dividing whole equation by 2

2x^2 + bx + 2ax + ab = 0

2x^2 + (2a+b)x + ab = 0

It is now a quadratic equation.

Apply quadratic formula:

{-b +- [ root(b^2-4ac)]}/ 2a

Here b= (2a+b), a = 2, c=ab

Substitute and solving we get,

[-(2a+b) +- 2a-b]/4 = values of x

=== -2a-b + 2a-b /4 ; ;; ;   -2a-b-2a+b / 4

= -2b/4 ; -4a/4

= -b/2 , -a

Hence the values of x can be -b/2 or -a