question_answer

If\[\frac{2x-3}{5}+\frac{x+3}{4}=\frac{4x+1}{7}\], find the value of x.

Answer:

Given,\[\frac{2x-3}{5}+\frac{x+3}{4}=\frac{4x+1}{7}\] \[\frac{4(2x-3)}{5\times 3}+\frac{5(x+3)}{4}=\frac{4x+l}{7}\] \[\frac{8x-12}{20}+\frac{5x+15}{20}=\frac{4x+1}{7}\] \[\frac{8x-12+5x+15}{20}=\frac{4x+1}{7}\] \[\frac{13x+3}{20}=\frac{4x+1}{7}\]                                                                                                                7(13x + 3) = 20(4x + 1) 91x + 21 = 80x + 20 91x - 80x =20-21 11x = -1 Thus,  x=\[-\frac{1}{11}\]