Question

Alguém consegue me ajudar com esses problemas do primeiro grau?$a) y- \frac{y}{3} = \frac{2}{3} b) x - \frac{5x}{2} = 1 \frac{1}{2} c) 3x - \frac{4x-1}{5} = 6 - \frac{2-5x}{6} d) 2( \frac{5-x}{3}) -6 ( \frac{x-1}{2} = \frac{x}{6} - 7 e) 3t [t-( \frac{t+2}{3} - t+2)} = 0$

Answer 1

a
y/1 - y/3 = 2/3
mmc = 3
( 3y - y) = 2
2y = 2
y = 2/2  = 1 ***
b
x1 - x/2 = 1 int 1/2
x/1 - x/2 = 3/2
mmc = 2
2x - x = 3
x = 3 ****
c
3x/1 - (4x - 1 )/5 = 6/1 - ( 2 - 5x)6
mmc = 30
90x - 6 ( 4x - 1) = 180 - 5 ( 2 - 5x )
90x - 24x + 6 = 180 - 10 + 25x
66x + 6   -180 + 10 - 25x = 0
41x  - 164 = 0
41x = 164
x = 164/4 = 41 ***
d
2 ( 5 - x)/3 - 6 ( x - 1)/5 = x/6 - 7/1
( 10 - 2x)/3 - ( 6x - 6)/5  = x/6 - 7/1
mmc = 30
10( 10 - 2x) - 6 ( 6x - 6) = 5x - 210
100 - 20x - 36x + 36 - 5x + 210 = 0
- 61x  + 346 = 0
61x - 346 = 0
61x = 346
x = 346/61 ***
e
3t/1 [ t/1 - ( t + 2)/3 - t/1 + 2/1 )]
mmc =3
9t[ 3t - 1 ( t + 2) - 3t + 6 ]
9t [ 3t - t - 2 - 3t + 6 ]
9t [ - t + 4 ]=
- 9t² + 36t = 
9t² - 36t = 0
t² - 4t = 0
t( t - 4) = 0
t = 0 ***
t - 4 = 0
t = 4 ****