Question

A) 2x-1supra 3+2x-1  supra4 + 2x-1 supra 12 =1

Answer 1

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

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

8x-4+6x-3+2x-1=12
8x+6x+2x=12+4+3+1
16x=20
x=$\frac{20}{16}$
x=$\frac{5}{4}$

Answer 2

( 2x - 1) / 3 + ( 2x - 1) / 4 + ( 2x - 1) / 12 = 1 ,    / = supra
numitorul comun = 12
4( 2x - 1) / 12 + 3( 2x - 1) / 12 + ( 2x - 1) / 12 = 12 /12
4( 2x - 1) + 3( 2x - 1) + ( 2x - 1) = 12
8x - 4 + 6x - 3 + 2x - 1 = 12
16x - 8 = 12
16x = 12 + 8
16x = 20
x = 20 / 16 , simplifici cu 4
x = 5 / 4