Question

x-3/x+3-x+3/x-3=48/7

Answer 1

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Answer 2

Answer:

$\text{The solutions are }\frac{9}{4}, -4$

Step-by-step explanation:

Given the equation

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

we have to solve the above equation.

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

$\frac{(x-3)(x-3)-(x+3)(x+3)}{(x+3)(x-3)}=\frac{48}{7}$

Opening brackets, we get

$\frac{x^2-6x+9-(x^2+6x+9)}{x^2-3^2}=\frac{48}{7}$   $(a^2-b^2=(a-b)(a+b))$

$\frac{-12x}{x^2-9}=\frac{48}{7}$

Cross multiplying both sides

$7(-12x)=48(x^2-9)$

$48x^2+84x-432=0$

$4x^2+7x-36=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-7\pm \sqrt{7^2-4(4)(-36)}}{2(4)}$

$x=\frac{-7\pm \sqrt{625}}{8}$

$x=\frac{-7+25}{8}, \frac{-7-25}{8}$

$x=\frac{18}{8}, \frac{-32}{8}$

$x=\frac{9}{4}, -4$

which are required solution.