Question

3.a. Solve the following equations.​

Answer 1

Answer:

1. Solving $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$ we get x=3

2. Solving $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$ we get x=-1

Step-by-step explanation:

We need to solve the equations:

1. $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$

Expanding the brackets

$\frac{2x}{3}-\frac{1}{3} -\frac{2x}{4}-\frac{1}{4} =\frac{2}{12}-\frac{x}{12} \\\frac{2x}{3}-\frac{1}{3} -\frac{x}{2}-\frac{1}{4} =\frac{1}{6}-\frac{x}{12}$

Taking LCM on left side we get 12

$\frac{2x*4-1*4-x*6-1*3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-4-6x-3}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{8x-6x-3-4}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x-7}{12}= \frac{1}{6}-\frac{x}{12}\\\frac{2x}{12}-\frac{7}{12}= \frac{1}{6}-\frac{x}{12}$

Add 7/12 on both sides

$\frac{2x}{12}-\frac{7}{12}+\frac{7}{12} = \frac{1}{6}-\frac{x}{12}+\frac{7}{12}\\\frac{x}{6}=\frac{1*2-x+7}{12}\\ \frac{x}{6}=\frac{-x+9}{12} \\\frac{x}{6}=-\frac{x}{12} +\frac{9}{12}\\ \frac{x}{6}=-\frac{x}{12} +\frac{3}{4}$

Adding x/12 on both sides and simplifying

$\frac{x}{6}+\frac{x}{12} =-\frac{x}{12} +\frac{3}{4}+\frac{x}{12}\\\frac{x*2+x}{12}=\frac{3}{4}\\\frac{3x}{12}= \frac{3}{4}$

Multiply both sides by 12/3

$\frac{3x}{12}*\frac{12}{3}=\frac{3}{4}*\frac{12}{3} \\Simplifying:\\x=3$

So, solving $\frac{1}{3}(2x-1)-\frac{1}{4}(2x+1)=\frac{1}{12}(2-x)$ we get x=3

2. $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$

Taking LCM and solving:

$\frac{x-2+2x-1}{4}=\frac{2x-1}{2}\\\frac{x+2x-1-2}{4}=\frac{2x-1}{2}\\\frac{3x-3}{4}=\frac{2x-1}{2}$

Cross multiplying

$2(3x-3)=4(2x-1)\\6x-6=8x-4\\Combining\ like\ terms \ :\\6x-8x=-4+6\\-2x=2\\x=-1$

So, solving $\frac{x-2}{4}+\frac{2x-1}{4}=x-\frac{1}{2}$ we get x=-1