question_answer

                                       Solve for x:\[\frac{3}{4}(7x-1)-\left[ 2x-\frac{1-x}{2} \right]=x+\frac{3}{2}\]

Answer:

\[\frac{21x}{4}-\frac{3}{4}-2x+\frac{1-x}{2}=x+\frac{3}{2}\]                                                                           \[\frac{21x}{4}-2x+\frac{1}{2}-\frac{x}{2}=x+\frac{3}{2}+\frac{3}{4}\] \[\frac{21x}{4}-2x+\frac{x}{2}-x=\frac{3}{2}+\frac{3}{4}-\frac{1}{2}\] \[\frac{21}{4}x-3x--\frac{x}{2}=\frac{3}{2}+\frac{3}{4}+\frac{1}{2}\] \[\frac{21x-4\times 3x-2\times x}{4}=\frac{6+3-1\times 2}{4}\] \[\frac{7x}{4}=\frac{7}{4}\] 7x =7 x =1