Solve the equation by factorisation:1/x-3-1/x+5=1/6​

Question

Solve the equation by factorisation:1/x-3-1/x+5=1/6​

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Aubrey 3 weeks 2021-11-07T09:31:44+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-07T09:33:27+00:00

    Answer:

    x = 7   OR  – 9

    Step-by-step explanation:

    Given equation :

    \sf\Rightarrow\dfrac{1}{x-3}-\dfrac{1}{x+5}  =\dfrac{1}{6}

    On simplifying the equation

    \sf\Rightarrow\dfrac{x+5-( x -3)}{(x-3)(x+5)} =\dfrac{1}{6}\\\\\\\sf\Rightarrow\dfrac{x+5-x+3}{x(x+5)-3(x+5)} =\dfrac{1}{6} \\\\\\\sf\Rightarrow\dfrac{8}{x^{2} +5x-3x-15} =\dfrac{1}{6 } \\\\\\ \sf\Rightarrow \dfrac{8}{x^{2} +2x-15}=\dfrac{1}{6}

    ⇒ 8( 6 ) = x² + 2x – 15

    ⇒ 48 = x² + 2x – 15

    ⇒ 0 = x² + 2x – 15 – 48

    ⇒ 0 = x² + 2x – 63

    Now solving by factorisation method

    ⇒ x² + 9x – 7x – 63 = 0

    ⇒ x( x + 9 ) – 7( x + 9 ) = 0

    ⇒ ( x – 7 )( x + 9 ) = 0

    ⇒ x – 7 = 0   OR   x + 9 = 0

    ⇒ x = 7   OR   x = – 9

    Hence the roots of the equation are 7 and – 9.

    0
    2021-11-07T09:33:34+00:00

    GIVEN EQUATION :-

    \sf{\dfrac{1}{x-3}-\dfrac{1}{x+5}  =\dfrac{1}{6}}

    TO FIND :-

    The values of x.

    SOLUTION :-

    \sf{\dfrac{1}{x-3}-\dfrac{1}{x+5}  =\dfrac{1}{6}}\\\\\\\sf{\longrightarrow \dfrac{(x+5)-(x-3)}{(x-3)(x+5)}=\dfrac{1}{6}}\\\\\\\sf{\longrightarrow \dfrac{x+5-x+3}{x^2+5x-3x-15}=\dfrac{1}{6}}\\\\\\\sf{\longrightarrow \dfrac{8}{x^2+2x-15}=\dfrac{1}{6}}

    By cross-multiplying –

    \sf{\longrightarrow x^2+2x-15=48}\\\\\sf{\longrightarrow x^2+2x-63=0}

    By middle-term factorising –

    \sf{\longrightarrow x^2+9x-7x-63=0}\\\\\sf{\longrightarrow x(x+9)-7(x+9)=0}\\\\\sf{\longrightarrow (x+9)(x-7)=0}\\\\\sf{(x+9)=0}\:\:\:\:\:\: \sf{or\:\:\:\:\:\: (x-7)=0}

    \Large\boxed{\sf{\implies x=-9}}\:\:\:\:\:\:\:\: \sf{or} \:\:\:\:\:\:\:\: \boxed{\sf{\implies x=7}}

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