Question

solve for x 1/x+1 + 3/5x+1 = 5/x+4​

Answer 1

$\dag\:\underline{\sf AnsWer :}$

$:\implies\sf \dfrac{1}{x+1}+\dfrac{3}{5x+1} = \dfrac{5}{x+4}$

$:\implies\sf \dfrac{1(5x + 1) + 3(x + 1)}{(x+1)(5x + 1)}= \dfrac{5}{x+4}$

$:\implies\sf \dfrac{5x + 1+ 3x + 3}{(x+1)(5x + 1)}= \dfrac{5}{x+4}$

$:\implies\sf \dfrac{8x +4}{ {5x}^{2} + x + 5x + 1}= \dfrac{5}{x + 4}$

$:\implies\sf \dfrac{8x +4}{ {5x}^{2} +6x + 1}= \dfrac{5}{x + 4}$

$:\implies\sf (8x +4)(x + 4)= 5( {5x}^{2} +6x + 1)$

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

$:\implies\sf {8x}^{2} + 36x + 16 = {25x}^{2} +30x + 5$

$:\implies\sf {25x}^{2} - {8x}^{2} +30x - 36x + 5 - 16 = 0$

$:\implies\sf {17x}^{2} - 6x - 11= 0$

$:\implies\sf {17x}^{2} - 17x + 11x - 11= 0$

$:\implies\sf 17x(x - 1) +11(x - 1)= 0$

$:\implies\sf (17x + 11)(x - 1)= 0$

$:\implies \underline{ \boxed{\sf x = \dfrac{ - 11}{17} \: or \: x = 1}}$

Answer 2

Correct Question-:

• Solve for x -: $\implies {\large{\sf{\frac{1}{x+1} +\frac{3}{5x+1} = \frac{5}{x+4}}}}$

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ The\:value\:of\:x\:is\:1\:or\:\frac{-11}{17} }}}}}$

$\dag{\sf{\large {EXPLANATION-:}}}$

• $\frak{Given \:\: -:} \begin{cases} \sf{Equation \: =\dfrac{1}{x+1} +\frac{3}{5x+1} = \frac{5}{x+4} \:}\end{cases}$

• $\frak{To \:Find\: -:} \begin{cases} \sf{Solve\: for \: x \:}\end{cases}$

$\dag{\sf{\large {Solution,}}}$

• $\implies {\large{\sf{\dfrac{1}{x+1} +\dfrac{3}{5x+1} = \dfrac{5}{x+4}}}}$

$\dag{\sf{\large {LCM = \:(x +1 )\:( 5x +1) }}}$

• $\implies {\large{\sf{\dfrac{1(5x+1)+3(x+1)}{(x+1)(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{5x+1+3x+3}{(x+1)(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{5x+3x+3+1}{(x+1)(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{8x+4}{(x+1)(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{8x+4}{x(5x+1)+1(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{8x+4}{5x^{2}+x+1(5x+1)} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{8x+4}{5x^{2}+x+5x+1} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{\dfrac{8x+4}{5x^{2}+6x+1} = \dfrac{5}{x+4}}}}$

• $\implies {\large{\sf{(8x+4)(x+4)=5(5x^{2}+6x+1)}}}$

• $\implies {\large{\sf{ 8x(x+4)+4(x+4)=5(5x^{2}+6x+1) }}}$

• $\implies {\large{\sf{ 8x^{2}+32x+4(x+4)=5(5x^{2}+6x+1) }}}$

• $\implies {\large{\sf{ 8x^{2}+32x+4x+16=5(5x^{2}+6x+1) }}}$

• $\implies {\large{\sf{ 8x^{2}+32x+4x+16=25x^{2}+30x+5 }}}$

• $\implies {\large{\sf{ 8x^{2}+36x+16=25x^{2}+30x+5 }}}$

$\dag{\sf{\large {By\:Shifting,}}}$

• $\implies {\large{\sf{ 25 x^{2} - 8x^{2}+36x-30x+16-5=0 }}}$

• $\implies {\large{\sf{ 17 x^{2} -6x-11=0 }}}$

$\dag{\sf{\large {By\:Expanding,}}}$

• $\implies {\large{\sf{ 17 x^{2} -17x+11x-11=0 }}}$

• $\implies {\large{\sf{ 17 x(x -1) +11(x-1)=0 }}}$

• $\implies {\large{\sf{ (17x +11) (x-1)=0 }}}$

$\dag{\sf{\large {Now,}}}$

• $\implies {\large{\sf{ 17x +11 =0 }}}$

• $\implies {\large{\sf{ 17x =-11 }}}$

• $\implies {\large{\sf{ x =\frac{-11}{17} }}}$

$\dag{\sf{\large {or,}}}$

• $\implies {\large{\sf{ x -1 =0 }}}$

• $\implies {\large{\sf{ x =1 }}}$

$\dag{\sf{\large {Hence,}}}$

• $\underline{\boxed{\star{\sf{\blue{ The\:value\:of\:x\:is\:1\:or\:\frac{-11}{17} }}}}}$

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