[tex]\Huge\bf\underline{\underline{\red{★Question:-}}}[/tex]
(x + 3)/(x – 2) – (1 – x)/x = 17/4
.

Pleas

[tex]\Huge\bf\underline{\underline{\red{★Question:-}}}[/tex]
(x + 3)/(x – 2) – (1 – x)/x = 17/4
.

Please solve.
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2 thoughts on “<br />[tex]\Huge\bf\underline{\underline{\red{★Question:-}}}[/tex]<br />(x + 3)/(x – 2) – (1 – x)/x = 17/4<br />.<br /><br />Pleas”

  1. ANSWER:

    Given:

    • (x + 3)/(x – 2) – (1 – x)/x = 17/4

    To Find:

    • Value of x

    Solution:

    We are given that,

    [tex]\implies\sf\dfrac{x+3}{x-2}-\dfrac{1-x}{x}=\dfrac{17}{4}[/tex]

    On taking LCM in LHS,

    [tex]\implies\sf\dfrac{(x+3)(x)-(1-x)(x-2)}{(x)(x-2)}=\dfrac{17}{4}[/tex]

    So,

    [tex]\implies\sf\dfrac{(x^2+3x)-(x-x^2-2+2x)}{x^2-2x}=\dfrac{17}{4}[/tex]

    On simplifying,

    [tex]\implies\sf\dfrac{x^2+3x-x+x^2+2-2x}{x^2-2x}=\dfrac{17}{4}[/tex]

    Solving like terms,

    [tex]\implies\sf\dfrac{2x^2+2}{x^2-2x}=\dfrac{17}{4}[/tex]

    On cross-multiplying,

    [tex]\implies\sf4(2x^2+2)=17(x^2-2x)[/tex]

    Hence,

    [tex]\implies\sf8x^2+8=17x^2-34x[/tex]

    Transposing LHS to RHS,

    [tex]\implies\sf0=17x^2-34x-8x^2-8 [/tex]

    Hence,

    [tex]\implies\sf9x^2-34x-8=0[/tex]

    On splitting the middle term,

    [tex]\implies\sf9x^2-36x+2x-8=0[/tex]

    Taking common,

    [tex]\implies\sf9x(x-4)+2(x-4)=0[/tex]

    Taking (x – 4) common,

    [tex]\implies\sf(x-4)(9x+2)=0[/tex]

    So,

    [tex]\implies\sf x-4=0\:\:\:and\:\:\:9x+2=0[/tex]

    Therefore,

    [tex]\implies\bf x=4\:\:\:and\:\:\:x=\dfrac{-2}{9}[/tex]

    Reply
  2. To find the answer of the given terms:-

    [tex]\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4} [/tex]

    [tex]x−2

    x+3

    x

    1−x

    =

    4

    17[/tex]

    Take LCM for the given terms:-

    [tex]\frac{(x+3)(x)-(x-2)(1-x)}{x(x-2)}=\frac{17}{4}

    x(x−2) \\

    (x+3)(x)−(x−2)(1−x)

    =

    4

    17[/tex]

    Cross multiply both left side and right side:-

    [tex]\begin{gathered}\begin{array}{l}{4\left(x^{2}+3 x-\left(x-2-x^{2}+2 x\right)\right)=17\left(x^{2}-2 x\right)} \\ {4\left(x^{2}+3 x+x^{2}-3 x+2\right)=17 x^{2}-34 x} \\ {4\left(2 x^{2}+2\right)=17 x^{2}-34 x}\end{array}\end{gathered}

    4(x

    2

    +3x−(x−2−x

    2

    +2x))=17(x

    2

    −2x)

    4(x

    2

    +3x+x

    2

    −3x+2)=17x

    2

    −34x

    4(2x

    2

    +2)=17x

    2

    −34x

    8 x^{2}+8=17 x^{2}-34 x8x

    2

    +8=17x

    2

    −34x[/tex]

    The quadratic equation is formed, find the roots of the quadratic equation.

    [tex]\begin{gathered}\begin{array}{l}{9 x^{2}-34 x-8=0} \\ {9 x^{2}-36 x+2 x-8=0} \\ {9 x(x-4)+2(x-4)=0}\end{array}\end{gathered}

    9x

    2

    −34x−8=0

    9x

    2

    −36x+2x−8=0

    9x(x−4)+2(x−4)=0

    [/tex]

    The value of x by solving the quadratic equation is:-

    [tex]\begin{gathered}\begin{array}{l}{(x-4)(9 x+2)=0} \\ {x=4,-\frac{2}{9}}\end{array}\end{gathered}

    (x−4)(9x+2)=0

    x=4,− 92

    [/tex]

    Therefore, x=4,

    [tex]\bold{\frac{-2}{9}} 9−2[/tex]

    [tex] if \frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}[/tex]

    Reply

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