Tentukan himpunan penyelesaian dari persamaan nilai mutlak A) |x+3|=2 B) |-6x+2|=4

A.) |x+3| = 2
☞ x+3 = 2
x = 2-3
x = -1
☞ -(x+3) = 2
-x-3 = 2
-x = 2+3
-x = 5
x = -5

HP : {-5 , -1}

B.) |-6x+2| = 4
☞ -6x+2 = 4
-6x = 4-2
-6x = 2
x = - 2/6
x = - 1/3
☞ -(-6x+2) = 4
6x-2 = 4
6x = 4+2
6x = 6
x = 1

HP : {- 1/3, 1}