The equations 2x+5= x+8 and 7x+7=6x+10 have same solution.

False

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True

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Solve the following system of equations in R.

$\frac{2 x + 1}{7 x - 1} > 5, \frac{x + 7}{x - 8} > 2$

\frac{x^{2} - 7 x + 10}{x^{2} - 5 x - 14} \times \frac{x^{3} + 8}{x^{3} - 8} \times \frac{x^{2} + 2 x + 4}{x^{2} - 2 x + 4} \times \frac{x - 7}{x - 5} = ?

\sqrt{2 x^{2} + 5 x - 7} + \sqrt{3 (x^{2} - 7 x + 6)} - \sqrt{7 x^{2} - 6 x - 1} = 0

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