tentukan himpunan penyelesaian dari |2x-1|=|x+4|

|2x-1| = |x+4|
(2x-1)² = (x+4)²
(2x-1)² - (x+4)² = 0
((2x-1) + (x+4))((2x-1) - (x+4)) = 0
(3x + 3)(x - 5) = 0
3x = -3 atau x = 5
x = -1

|2x+4|=|x+4|
(
[tex] (\sqrt{ 2x + 4 { }^{2} })^{2} = (\sqrt{x + 4 {}^{2} })^{2} \\ 2x + 4 {}^{2} = x + 4 {}^{2} \\ 4x {}^{2} + 16x + 16 = x {}^{2} + 8x + 16 \\ 4x {}^{2} - x {}^{2} + 16x - 8x + 16 - 16 = 0 \\ 3x {}^{2} + 8x = 0 \\ x(3x + 8) = 0 \\ x = 0 \: atau \: x = - 8 \div 3[/tex]