Pomoże ktoś. Zadanie w załączniku

[tex]a)\\\\\frac{2x+3}{5}=\frac{2-x}{3}-x\ \ /\cdot15\\\\3(2x+3)=5(2-x)-15x\\\\6x+9=10-5x-15x\\\\6x+9=10-20x\\\\6x+20x=10-9\\\\26x=1\ \ /:26\\\\x=\frac{1}{26}\\\\\\b)\\\\\frac{x+1}{2}-\frac{x-1}{3}=\frac{1}{4}\ \ /\cdot12\\\\6(x+1)-4(x-1)=3\\\\6x+6-4x+4=3\\\\2x+10=3\\\\2x=3-10\\\\2x=-7\ \ /:2\\\\x=-\frac{7}{2}\\\\x=-3\frac{1}{2}[/tex]

[tex]c)\\\\\frac{4x+7}{2}-\frac{5x-1}{6}=x+5\ \ /\cdot6\\\\3(4x+7)-(5x-1)=6x+30\\\\12x+21-5x+1=6x+30\\\\7x+22=6x+30\\\\7x-6x=30-22\\\\x=8\\\\\\d)\\\\\frac{x+5}{4}-\frac{2x+1}{8}=\frac{1}{2}(2-x)\\\\\frac{x+5}{4}-\frac{2x+1}{8}=1-\frac{1}{2}x\ \ /\cdot8\\\\2(x+5)-(2x+1)=8-4x\\\\2x+10-2x-1=8-4x\\\\9=8-4x\\\\4x=8-9\\\\4x=-1\ \ /:4\\\\x=-\frac{1}{4}[/tex]

LoczeeQviz LoczeeQviz · Answer 2

a)

[tex] \frac{2x + 3}{5} = \frac{2 - x}{3} - x | \times 15 \\ 3(2x + 3) = 5(2 - x) - 15x \\ 6x + 9 = 10 - 5x - 15x \\ 6x + 20x = 1 \\ 26x = 1| \div 26 \\ x = \frac{1}{26} [/tex]

b)

[tex] \frac{x + 1}{2} - \frac{x - 1}{3} = \frac{1}{4} | \times 12 \\ 6(x + 1) - 4(x - 1) = 3 \\ 6x + 6 - 4x + 4 = 3 \\ 2x = 3 - 10 \\ 2x = - 7 | \div 2 \\ x - \frac{7}{2} \\ x = - 3.5[/tex]

c)

[tex] \frac{4x + 7}{2} - \frac{5x - 1}{6} = x + 5 | \times 6 \\ 3(4x + 7) - 1(5x - 1) = 6(x + 5) \\ 12x + 21 - 5x + 1 = 6x + 30 \\ 7x - 6x = 30 - 22 \\ x = 8[/tex]

d)

[tex] \frac{x + 5}{4} - \frac{2x + 1}{8} = \frac{1}{2} (2 - x) | \times 8 \\ 2(x + 5) - 1(2x + 1) = 4(2 - x) \\ 2x + 10 - 2x - 1 = 8 - 4x \\ 4x = - 1 | \div 4 \\ x = - \frac{1}{4} [/tex]