abs(2x+1)/3+4=10

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Solution for abs(2x+1)/3+4=10 equation:


x in (-oo:+oo)

abs(2*x+1)/3+4 = 10 // - 10

abs(2*x+1)/3-10+4 = 0

abs(2*x+1)/3-10+4 = 0

x in (-oo:+oo)

abs(2*x+1)

2*x+1 >= 0

2*x+1 >= 0 // - 1

2*x >= -1 // : 2

x >= -1/2

x in <-1/2:+oo)

(2*x+1)/3-10+4 = 0

x in (-oo:-1/2)

(-(2*x+1))/3-10+4 = 0

abs(2*x+1)/3-10+4 = 0

/| (-(2*x+1))/3-10+4 = 0 i x in (-oo:-1/2)| (2*x+1)/3-10+4 = 0 i x in <-1/2:+oo)

x in (-oo:-1/2)

(-(2*x+1))/3-10+4 = 0

(-2*x-1)/3-10+4 = 0

(-2*x-1)/3+(-10*3)/3+(3*4)/3 = 0

3*4-2*x-10*3-1 = 0

12-2*x-31 = 0

-2*x-19 = 0

(-2*x-19)/3 = 0

(-2*x-19)/3 = 0 // * 3

-2*x-19 = 0

-2*x-19 = 0 // + 19

-2*x = 19 // : -2

x = 19/(-2)

x = -19/2

x in <-1/2:+oo)

(2*x+1)/3-10+4 = 0

(2*x+1)/3+(-10*3)/3+(3*4)/3 = 0

2*x-10*3+3*4+1 = 0

2*x-29+12 = 0

2*x-17 = 0

(2*x-17)/3 = 0

(2*x-17)/3 = 0 // * 3

2*x-17 = 0

2*x-17 = 0 // + 17

2*x = 17 // : 2

x = 17/2

x in { -19/2, 17/2 }

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