Respuesta :
Explanation:
[tex]\frac{3}{4}\mleft(6x+1\mright)-3x=\frac{1}{4}\mleft(2x-1\mright)[/tex]We expand the parenthesis using distributive property:
[tex]\begin{gathered} \frac{3}{4}(6x)\text{ + }\frac{3}{4}(1)\text{ - 3x = }\frac{1}{4}(2x)\text{ - }\frac{1}{4}(1) \\ \frac{18x}{4}\text{ + }\frac{3}{4}\text{ - 3x = }\frac{2x}{4}-\text{ }\frac{1}{4} \\ \\ \text{collect like terms:} \\ \frac{18x}{4}\text{ - 3x - }\frac{2x}{4}\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \end{gathered}[/tex][tex]\begin{gathered} \frac{18x}{4}\text{ -}\frac{\text{ 3x}}{1}\text{ - }\frac{2x}{4}\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \\ we\text{ find the LCM of left hand side:} \\ \frac{18x\text{ -3x(4) - 2x}}{4}\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \\ \frac{18x\text{ - 12x -2x}}{4}\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \\ \frac{4x}{4}\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \\ x\text{ = }-\text{ }\frac{1}{4}\text{ -}\frac{3}{4}\text{ } \end{gathered}[/tex][tex]undefined[/tex]