Question

Please solve this fast. Right answer will be marked as brainliest. Donot spam.

Answer 1

$\LARGE{ \underline{ \green{ \sf{Required \: answer:}}}}$

We are given with an equation where we can see the variable is x. And we have to solve for the value of x.

GiveN:

• $\rm{ \dfrac{x - 1}{x + 2} + \dfrac{x - 3}{x} = \dfrac{29}{20} }$

Taking LCM in the left hand side of the eq.

⇛ $\rm{\dfrac{x(x - 1) + (x - 3)(x + 2)}{x(x + 2)} = \dfrac{29}{20} }$

Expanding the parentheses in the num. and den.

⇛ $\rm{ \dfrac{ {x}^{2} - x + {x}^{2} -3x + 2x - 6 }{ {x}^{2} + 2x } = \dfrac{29}{20} }$

Combining the like terms in the num.

⇛ $\rm{ \dfrac{2 {x}^{2} - 2x - 6}{ {x}^{2} + 2x } = \dfrac{29}{20} }$

Cross multiplying,

⇛ $\rm{20(2 {x}^{2} - 2x - 6) = 29( {x}^{2} + 2x)}$

Expanding the parentheses,

⇛ $\rm{40 {x}^{2} - 40x - 120 = 29 {x}^{2} + 58x}$

⇛ $\rm{40 {x}^{2} - 40x - 120 - 29 {x}^{2} - 58x = 0}$

Combining the like terms again,

⇛ $\rm{11 {x}^{2} - 98x - 120 = 0}$

Finding zeroes by using middle term factorisation,

⇛ $\rm{11 {x}^{2} - 110x + 12x - 120 = 0}$

⇛ $\rm{11x(x - 10) + 12(x - 10) = 0}$

⇛ $\rm{(11x + 12)(x - 10) = 0}$

This means, when each of them is equated to 0, the value of x will be -12/11 or 10 (Answer)

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

$\sf{Solve\:for\:x\:\:in:\:\dfrac{x-1}{x+2}+\dfrac{x-3}{x}=\dfrac{29}{20}}$

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{x=10,\:x=-\dfrac{12}{11}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\text { Find Least Common Multiplier of } x+2, x, 20$

$\bullet\:\:\mathrm{The\:LCM\:of\:}a,\:b\:\mathrm{is\:the\:smallest\:multiplier\:that\:is\:divisible\:by\:both\:}a\mathrm{\:and\:}b \bullet\:\:$

Compute an expression comprised of factors that appear in at least one of the factored expressions

$=20x\left(x+2\right)$

$\mathrm{Multiply\:by\:LCM=}20x\left(x+2\right)$

$\dfrac{x-1}{x+2}\cdot \:20x\left(x+2\right)+\dfrac{x-3}{x}\cdot \:20x\left(x+2\right)=\dfrac{29}{20}\cdot \:20x\left(x+2\right)$

$20x\left(x-1\right)+20\left(x-3\right)\left(x+2\right)=29x\left(x+2\right)$

$\begin{array}{l}\text { Expand } 20 x(x-1)+20(x-3)(x+2): 40 x^{2}-40 x-120 \\\\\text { Expand } 29 x(x+2): \quad 29 x^{2}+58 x\end{array}$

$40x^2-40x-120=29x^2+58x$

$\implies40x^2-40x-120-58x=29x^2+58x-58x$

$\implies40x^2-98x-120=29x^2$

$\implies40x^2-98x-120-29x^2=29x^2-29x^2$

$\implies11x^2-98x-120=0$

Solve with the quadratic formula

Quadratic Equation Formula:

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\mathrm{For\:}\quad a=11,\:b=-98,\:c=-120:\quad x_{1,\:2}=\dfrac{-\left(-98\right)\pm \sqrt{\left(-98\right)^2-4\cdot \:11\left(-120\right)}}{2\cdot \:11}$

$\begin{array}{l}x=\dfrac{-(-98)+\sqrt{(-98)^{2}-4 \cdot 11(-120)}}{2 \cdot 11}: 10 \\\\\\x=\dfrac{-(-98)-\sqrt{(-98)^{2}-4 \cdot 11(-120)}}{2 \cdot 11}:-\dfrac{12}{11}\end{array}$

$\large\boxed{\sf{x=10,\:x=-\dfrac{12}{11}}}$