Question

solve for x : 1/x+1 + 3/ 5x+1 = 5/x+4 where x is not equal to - 1, - 1/5, - 4... see attachment for convenience.... plzz solve... ​

Answer 1

$\dfrac{1}{x + 1} + \dfrac{3}{5x + 1} = \dfrac{5}{x + 4}$, x ≠ -1, -$\dfrac{1}{5}$, -4

Move the expression to the left

☛ $\dfrac{1}{x + 1} + \dfrac{3}{5x + 1} - \dfrac{5}{x + 4}$

Write all numerators above the common denominators

☛ $\dfrac{(5x + 1) \times (x + 4) + 3(x + 1) \times (x + 4) - 5(x + 1) \times (5x + 1}{(x + 1) \times (5x + 1) \times (x + 4)}$

Remove the brackets

☛ $\dfrac{5x^2 + 20x + x + 4 + (3x + 3) \times (x + 4) + (-5x - 5) \times (5x + 1)}{(x + 1) \times (5x + 1) \times (x + 4)}\:=\:0$

Multiple the brackets

☛ $\dfrac{5x^2 + 20x + x + 4 + 3x^2 + 12x + 3x + 12 - 25x^2 - 5x - 25x - 5}{(x + 1) \times (5x + 1) \times (x + 4)}\:=\:0$

Collect the like terms and calculate

☛ $\dfrac{-17x^2 + 6x + 11}{(x + 1) \times (5x + 1) \times (x + 4)}\:=\:0$

Set the numerator equal to 0

☛ $-17x^2 + 6x + 11 = 0$

Change the signs

☛ $17x^2 - 6x - 11 = 0$

Solve the quadriatic equation

☛ $x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 17 \times (-11)}}{2 \times 17}$

Remove brackets and calculate

☛ $x = \dfrac{6 \pm \sqrt{36 + 748}}{34}$

Add the numbers

☛ $x = \dfrac{6 \pm \sqrt{784}}{34}$

Calculate the root

☛ $x = \dfrac{6 \pm 28}{34}$

Sepatate the solutions

☛ $x = \dfrac{6 + 28}{34}$

$x = \dfrac{6 - 28}{34}$

Simplify the expressions

☛ $x = 1$

$x = -\dfrac{11}{17}$

x ≠ -1, -$\dfrac{1}{5}$, -4

Check if the solution is in the defined range

☛ $x = 1$

$x = -\dfrac{11}{17}$

The final solutions are :-

☛ $x_1 = 1,\: x_2 = -\dfrac{11}{17}$

Answer 2

Refer to the attachment....

Hope it helps...

ples mark brainliest... ☺