Question

2/3x+1/2(x-5)=3/2(x-1)​

Answer 1

Answer:

$x = \frac{ - 6 + \sqrt{132} }{12} \\ \\ x = \frac{ - 6 - \sqrt{132} }{12}$

Step-by-step explanation:

$\frac{2}{3x} + \frac{1}{2}(x - 5) = \frac{3}{2} (x - 1) \\ \\ = > \frac{2}{3x} + \frac{x - 5}{2} = \frac{3x - 3}{2} \\ \\ = > \frac{(2 \times 2) + 3x(x - 5)}{3x \: \times 2} = \frac{3x - 3}{2} \\ \\ = > \frac{4 + {3x}^{2} - 15x }{ \cancel 6x} = \frac{3x - 3}{ \cancel 2} \\ \\ = > \frac{4 + {3x}^{2} - 15x}{3x} = \frac{3x - 3}{1} \\ \\ = > 4 + 3 {x}^{2} - 15x = 3x(3x - 3) \\ \\ = > 4 + 3 {x}^{2} - 15x =9 {x}^{2} - 9x \\ \\ = > 0 = 9 {x}^{2} - 3 {x}^{2} - 9x + 15x - 4 \\ \\ = > 0 = 6 {x}^{2} + 6x - 4 \\ \\ Using \: quadratic \: formula \\ \\ Here \\ a = 6 \\ b = 6 \\ c = - 4 \\ \\ Quadratic \: formula \: = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \sqrt{ {b}^{2} - 4ac} \\ = \sqrt{ {6}^{2} - (4 \times 6 \times - 4) } \\ \\ = \sqrt{36 - ( - 96)} \\ = \sqrt{36 + 96} \\ = \sqrt{132} \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ - 6 + \sqrt{132} }{2 \times 6} \\ \\ x = \frac{ - 6 + \sqrt{132} }{12} \\ \\ \\ \\ \\ x = \frac{ - 6 - \sqrt{132} }{6 \times 2} \\ x = \frac{ - 6 - \sqrt{132} }{12}$

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\displaystyle \: \frac{2}{3x} + \frac{1}{2(x - 5)} = \frac{3}{2(x - 1)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{3}{2(x - 1)} - \frac{1}{2(x - 5)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{3(x - 5) - (x - 1)}{2(x - 1)(x - 5)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{(3x - 15 - x + 1)}{2(x - 1)(x - 5)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{(2x - 14)}{2(x - 1)(x - 5)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{2(x - 7)}{2(x - 1)(x - 5)}$

$\implies \: \displaystyle \: \frac{2}{3x} = \frac{(x - 7)}{(x - 1)(x - 5)}$

$\implies \: \displaystyle \: 2(x - 1)(x - 5) = 3x(x - 7)$

$\implies \: \displaystyle \: 2( {x}^{2} - 6x + 5 ) = 3( {x}^{2} - 7x)$

$\implies \: \displaystyle \: 2 {x}^{2} - 12x + 10 = 3 {x}^{2} - 21x$

$\implies \: \displaystyle {x}^{2} - 9x - 10 = 0$

$\implies \: \displaystyle {x}^{2} - 10x + x - 10 = 0$

$\implies \: \displaystyle x(x - 10) + 1(x - 10) = 0$

$\implies \: \displaystyle (x - 10)(x + 1) = 0$

We know that if the product of two real numbers are zero then either of them are zero

So

$either \: \: x - 10 = 0 \: \: \: or \: \: x + 1 = 0$

Now

$x - 10 = 0 \: \: gives \: \: x = 10$

And

$x + 1 = 0 \: \: \: gives \: \: \: x = - 1$

RESULT

$\red{\fbox{THE \: REQUIRED \: SOLUTION \: IS \: x = - 1 , 10}}$