Solve for x: $\frac{1}{x+1} +\frac{3}{5x+1} =\frac{5}{x+4} ,\ x
eq 1,\ -\frac{1}{5} ,\ -4$


Given: Equation $\frac{1}{x+1} +\frac{3}{5x+1} =\frac{5}{x+4} ,\ x\
eq 1,\ -\frac{1}{5} ,\ -4$

To do: To find the value of x by solving the given equation.

Solution:

The given equation: $\frac{1}{x+1} +\frac{3}{5x+1} =\frac{5}{x+4}$

$\Rightarrow \frac{5x+1+3x+3}{\left( x+1\right)\left( 5x+1\right)} =\frac{5}{x+4}$

$\Rightarrow \frac{6x+2}{5x^{2} +5x+x+1} =\frac{5}{x+4}$

$\Rightarrow \left( 8x+4\right)\left( x+4\right) =5\left( 5x^{2} +6x+1\right)$

$\Rightarrow 8x^{2} +32x+4x+16=25x^{2} +30x+5$

$\Rightarrow 25x^{2} +30x+5-8x^{2} -32x-4x-16=0$

$\Rightarrow 17x^{2} -6x-11=0$

$\Rightarrow 17x^{2} -17x+11x-11=0$

$\Rightarrow 17x\left( x-1\right) +11\left( x-1\right) =0$

$\Rightarrow ( 17x+11)( x-1) =0$

If $17x+11=0$

$\Rightarrow x=-\frac{11}{17}$

If $x-1=0$

$\Rightarrow x=1$

$\therefore x=-\frac{11}{17} ,\ 1$.

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Updated on: 10-Oct-2022

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