Question

помогитеееееее с решением пожалуйста​

Answer 1

$\dfrac{x+1}{2}=\dfrac{2x-3}{5}+2\\\\\dfrac{x+1}{2}\cdot10=\dfrac{2x-3}{5}\cdot10 + 2\cdot10\\\\(x+1)\cdot5 = (2x-3)\cdot2 + 2\cdot10\\\\5x + 5 = 4x - 6 + 20\\\\5x - 4x = 20 - 6 - 5\\\\x = 9$

$\dfrac{11-3x}{4} = \dfrac{x+13}{6} - x\\\\\dfrac{11-3x}{4}\cdot12 = \dfrac{x+13}{6}\cdot12 - x\cdot12\\\\(11-3x)\cdot3 = (x+13)\cdot2 - x\cdot12\\\\33 - 9x = 2x + 26 - 12x\\\\12x - 2x - 9x = 26 - 33\\\\x = -7$

$\dfrac{5x+8}{9} + x = \dfrac{7x+10}{6}\\\\\dfrac{5x+8}{9}\cdot18 + x\cdot18 = \dfrac{7x+10}{6}\cdot18\\\\(5x+8)\cdot2 + x\cdot18 = (7x+10)\cdot3\\\\10x+16+18x = 21x+30\\\\10x+18x-21x = 30-16\\\\7x = 14\\\\x = 2$

$\dfrac{2\cdot(2-3x)}{5} + 1 = \dfrac{3-4x}{3}\\\\\dfrac{2\cdot(2-3x)}{5}\cdot15 + 1\cdot15 = \dfrac{3-4x}{3}\cdot15\\\\2\cdot(2-3x)\cdot3 + 15 = (3-4x)\cdot5\\\\12 - 18x + 15 = 15 - 20x\\\\20x - 18x = 15 - 15 - 12\\\\2x = -12\\\\x=-6$