Question

$\frac{3}{2x - 1} + \frac{4}{2x + 1} = \frac{7}{2x}$

Find out the value of x​

Answer 1

$= > \frac{3}{2} x - 1 + \frac{4}{2} x + 1 = \frac{7}{2} x$

$= > \frac{3x}{2} - 1 + \frac{4}{2} x + 1 = \frac{7}{2}x$

$= > \frac{3x}{2} - 1 + 2x + 1 = \frac{7}{2}x$

$= > \frac{3x}{2} + 2x = \frac{7}{2} x$

$= > \frac{3x}{2} + 2x = \frac{7x}{2}$

$= > 2 \frac{3x}{2} + 2.2x = 2 \frac{7x}{2}$

$= > 3x + 2.2x = 7x$

$= > 3x + 4x = 7x$

$= > 7x = 7x$

$= > 7x - 7x = 7x - 7x$

$= > 0 = 0$

I hope this will help you dear..

Always stay safe and stay healthy..

Answer 2

Solution:

$\sf{\dfrac{3}{2x-1}+\dfrac{4}{2x+1}=\dfrac{7}{2x}}$

Taking LCM,

$\Rightarrow \: \: \sf{\dfrac{3(2x+1)+4(2x-1)}{(2x-1)(2x+1)}=\dfrac{7}{2x}}$

$\Rightarrow \ \sf{\dfrac{6x+3+8x-4}{4x^{2} -1}=\dfrac{7}{2x}}$

Further simplifying,

$\Rightarrow \: \sf{\dfrac{14x-1}{4x^{2} -1}=\dfrac{7}{2x}}$

On cross multiplication,

$\Rightarrow \: \sf{7(4x^{2} -1)=2x(14x-1)}$

$\Rightarrow \: \sf{28x^{2} -7=28x^{2} -2x}$

$\Rightarrow \: \sf{-7=-2x}$

$\Rightarrow \: \sf{2x=7}$

$\Rightarrow \: \sf{x=\dfrac{7}{2}}$

$\Rightarrow \: \sf{x=3.5}$

$\therefore \boxed{\bf{x=3.5}}$

KNOW MORE:

• (a + b)² = a² + b² + 2ab
• (a - b)² = a² + b² - 2ab
• a² - b² = (a + b)(a - b)
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ac)
• (a + b)³ = a³ + b³ + 3ab(a + b)