Answer:
1. Solving [tex]\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)[/tex] we get x=3
2. Solving [tex]\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}[/tex] we get x=-1
Step-by-step explanation:
We need to solve the equations:
1. [tex]\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)[/tex]
Expanding the brackets
[tex]\frac{2x}{3}-\frac{1}{3} -\frac{2x}{4}-\frac{1}{4} =\frac{2}{12}-\frac{x}{12} \\\frac{2x}{3}-\frac{1}{3} -\frac{x}{2}-\frac{1}{4} =\frac{1}{6}-\frac{x}{12}[/tex]
Taking LCM on left side we get 12
[tex]\frac{2x*4-1*4-x*6-1*3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-4-6x-3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-6x-3-4}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x-7}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x}{12}-\frac{7}{12}= \frac{1}{6}-\frac{x}{12}[/tex]
Add 7/12 on both sides
[tex]\frac{2x}{12}-\frac{7}{12}+\frac{7}{12} = \frac{1}{6}-\frac{x}{12}+\frac{7}{12}\\\frac{x}{6}=\frac{1*2-x+7}{12}\\ \frac{x}{6}=\frac{-x+9}{12} \\\frac{x}{6}=-\frac{x}{12} +\frac{9}{12}\\ \frac{x}{6}=-\frac{x}{12} +\frac{3}{4}[/tex]
Adding x/12 on both sides and simplifying
[tex]\frac{x}{6}+\frac{x}{12} =-\frac{x}{12} +\frac{3}{4}+\frac{x}{12}\\\frac{x*2+x}{12}=\frac{3}{4}\\\frac{3x}{12}= \frac{3}{4}\\[/tex]
Multiply both sides by 12/3
[tex]\frac{3x}{12}*\frac{12}{3}=\frac{3}{4}*\frac{12}{3} \\Simplifying:\\x=3[/tex]
So, solving [tex]\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)[/tex] we get x=3
2. [tex]\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}[/tex]
Taking LCM and solving:
[tex]\frac{x-2+2x-1}{4}=\frac{2x-1}{2}\\\frac{x+2x-1-2}{4}=\frac{2x-1}{2}\\\frac{3x-3}{4}=\frac{2x-1}{2}\\[/tex]
Cross multiplying
[tex]2(3x-3)=4(2x-1)\\6x-6=8x-4\\Combining\ like\ terms \ :\\6x-8x=-4+6\\-2x=2\\x=-1[/tex]
So, solving [tex]\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}[/tex] we get x=-1
