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