We can use this same process if there are parenthesis in the problem. We will first distribute the coefficient in front of the parenthesis, then clear the fractions. This is seen in the following example. Example `3/2(5/9x+4/27)=3`

`3/2(5/9x+4/27)=3`	Distribute `3/2` through parenthesis, reducing if possible
`5/6 x + 2/9 = 3`	`LCD = 18`, multiply each term by `18`
`color(red)(18)5/6 x + color(red)(18)2/9 = color(red)(18)(3)`	Reduce `18` with each denominator
`(color(red)3)(5x) + color(red)(2)(2) = color(red)(18)(3)`	Multiply out each term
`15x+4=54`	Focus on addition of `4`
`15x+4color(red)(-4)=54color(red)(-4)`	Subtract 4 from both sides
`15x=50`	Focus on multiplication by 15
`(15x)/color(red)(15)=50/color(red)(15)`	Divide both sides by 15. Reduce on right side
`x= 10/3`	Solution

While the problem can take many different forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term by the LCD and reduce. However, you may consider this step is necessary or not. Example `3/4x− 1/2 = 1/3(3/4x+6)− 7/2`

`3/4x− 1/2 = 1/3(3/4x+6)− 7/2`	Distribute 1/3 , reduce if possible
`3/4x− 1/2 = 1/4x+2− 7/2`	instant of find LCD, let as focus on `1/2` first
`3/4x− 1/2color(red)(+1/2) = 1/4x+2− 7/2color(red)(+1/2)`	Add `1/2` on both side
`3/4x= 1/4x+2− 3`	Not necessary find LCD now
`3/4x= 1/4x-1`	Focus on `1/4 x`
`3/4xcolor(red)(-1/4x)= 1/4xcolor(red)(-1/4x)-1`	Subtract `1/4x` from both sides
`1/2x=-1 `	Focus on multiplication by `1/2 `
`color(red)((2))1/2x=-1color(red)((2))`	multiply `2` on both sides
`x= -2`	Solution