abs((3x-2)/2)5

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Solution for abs((3x-2)/2)5 equation:


x in (-oo:+oo)

abs((3*x-2)/2) < 5 // - 5

abs((3*x-2)/2)-5 < 0

abs((3*x-2)/2)-5 < 0

x in (-oo:+oo)

abs((3*x-2)/2)

(3*x-2)/2 >= 0

( 2 )

2 >= 0

x należy do R

( 3*x-2 )

3*x-2 >= 0 // + 2

3*x >= 2 // : 3

x >= 2/3

(-oo:2/3) (2/3:+oo) 2 + + 3*x-2 - +

x in <2/3:+oo)

(3*x-2)/2-5 < 0

x in (-oo:2/3)

-((3*x-2)/2)-5 < 0

abs((3*x-2)/2)-5 < 0

/ | -((3*x-2)/2)-5 < 0 i x in (-oo:2/3) | (3*x-2)/2-5 < 0 i x in <2/3:+oo)

x in (-oo:2/3)

-((3*x-2)/2)-5 < 0

(-1*(3*x-2))/2-5 < 0

(-1*(3*x-2))/2+(-5*2)/2 < 0

-1*(3*x-2)-5*2 < 0

-3*x-8 < 0

(-3*x-8)/2 < 0

(-3*x-8)/2 < 0 // * 2^2

2*(-3*x-8) < 0

( -3*x-8 )

-3*x-8 < 0 // + 8

-3*x < 8 // : -3

x > 8/(-3)

x > -8/3

( 2 )

2 < 0

x belongs to the empty set

(-oo:-8/3) (-8/3:+oo) -3*x-8 + - 2 + +

x in <2/3:+oo)

(3*x-2)/2-5 < 0

(3*x-2)/2+(-5*2)/2 < 0

3*x-5*2-2 < 0

3*x-12 < 0

(3*x-12)/2 < 0

(3*x-12)/2 < 0 // * 2^2

2*(3*x-12) < 0

( 2 )

2 < 0

x belongs to the empty set

( 3*x-12 )

3*x-12 < 0 // + 12

3*x < 12 // : 3

x < 12/3

x < 4

(-oo:4) (4:+oo) 2 + + 3*x-12 - +

x in (-8/3:4)

(-8/3:4)

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