Question

1/x + 1/x+2 = 12/35​

Answer 1

$\red{ \huge{ \underline{ \underline{ \overline{ \mid{ \bold{ \: \:SOLUTION \: { \mid}}}}}}}}$

$\bold{ \frac{1}{x} + \frac{1}{ x + 2} = \frac{12}{35} } \\ \\ \bold{ = > \frac{(x + 2) + x}{x(x + 2)} = \frac{12}{35} } \\ \\ \bold{ = > \frac{2x + 2}{x {}^{2} + 2x } = \frac{12}{35} } \\ \\ \\ \bold{\underline{ By \: \: cross \: \: multiplcation}} \\ \\ \bold{ \frac{2x + 2}{x {}^{2} + 2x} = \frac{12}{35} } \\ \\ \bold{ = > 12(x {}^{2} + 2x) = 35(2x + 2) } \\ \\ \bold{ = > 12x {}^{2} + 24x = 70x + 70} \\ \\ \bold{ = > 12x {}^{2} + 24x - 70x - 70 = 0} \\ \\ = > \bold{12x {}^{2} - 46x - 70 = 0} \\ \\ \bold{=>6x {}^{2} - 23x - 35 = 0 }$

$\bold{\underline{Spliting \: \: the \: \: middle \: \: term \: \: }} \\ \\ \bold{{6x }^{2} - 23x - 35 = 0 } \\ \\ = > \bold{6x {}^{2} - 30x + 7x - 35 = 0} \\ \\ \bold{ = > 6x(x - 5) + 7(x - 5) = 0} \\ \\ \bold{ = > (x - 5)(6x + 7) = 0}$

$\bold{Now,} \\ \\ \bold{x - 5 = 0 \: \: \: \: \: Or, \: \: \: \: \: \: 6x + 7 = 0} \\ \\ \bold{ = > x = 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 6x = - 7} \\ \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x = - \: \: \frac{ 7}{6} }$

$\red{ \huge{ \underline{ \underline{ \overline{ \mid{ \bold{ \: \: VERIFICATION \: { \mid}}}}}}}}$

$\bold{Putting \: \: the \: \:value \: \: of \: \: x = 5 \: \: in \: \: \: L.H.S.} \\ \\ \bold{ L.H.S. = \frac{1}{x} + \frac{1}{x + 2} } \\ \\ \bold{ = \frac{1}{5} + \frac{1}{5 + 2} } \\ \\ \bold{ = \frac{1}{5} + \frac{1}{7} } \\ \\ = \bold{ = \frac{7 +5 }{35} } \\ \\ \bold{ = \frac{12}{35} } \\ \\ \bold{ = R.H.S.}$

$\bold{Again} \\ \\ \bold{Putting \: \: the \: \: value \: \: of \: \: x = - \: \frac{7}{6} \: \: in \: \: L.H.S.} \\ \\ \bold{L.H.S.= \frac{1}{x} + \frac{1}{x + 2} } \\ \\ \bold{ = \frac{1}{( - \frac{7}{6} )} + \frac{1}{( - \frac{7}{6} ) + 2} } \\ \\ \bold{ = - \frac{6}{7} + \frac{ 6}{5} } \\ \\ \bold{ = \frac{ - 30 + 42}{35} } \\ \\ \bold{ = \frac{12}{35} } \\ \\ \bold{ = R.H.S.}$

$\: \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \: \: \: Hence \: \: Verified. \: \: \: }}$

$\red{ \huge{ \underline{ \underline{ \overline{ \mid{ \bold{ \: \: ANSWER\: { \mid}}}}}}}}$ ➡

$\boxed { \boxed{ \purple{ \bold{ \: \: x = 5 \: \: \: \: \: \: \: Or \: \: \: \: \: \: \:x= - \: \frac{7}{6} \: \: }}}}$

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✝$\mathfrak{\huge{ \red{ \: THANKS..}}}$✝

Answer 2

Answer: x = 0 or x = -7/6

Step-by-step explanation:

Given,

$\frac{1}{x}+\frac{1}{x+2}=\frac{12}{35}\\\;\\\text{Taking LCM,}\\\;\\\frac{x+2+x}{x(x+2)}=\frac{12}{35}\\\;\\\frac{2x+2}{x(x+2)}=\frac{12}{35}\\\;\\\frac{2(x+1)}{x^2+2x}=\frac{12}{35}\\\;\\\frac{x+1}{x^2+2x}=\frac{6}{35}\\\;\\35(x+1)=6(x^2+2x)\\\;\\35x+35=6x^2+12x\\\;\\6x^2+12x-35x-35=0\\\;\\6x^2-23x-35=0\\\;\\6x^2-30x+7x-35=0\\\;\\6x(x-5)+7(x-5)=0\\\;\\(x-5)(6x+7)=0\\\;\\x=5\;\;or\;\;\;x=-\frac{7}{6}$