How do you solve #4 |4-3x |= 4x + 6# and find any extraneous solutions?

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Answer 1 · 2016-06-28T14:27:47

#x=5/8 and x=11/4#

The equation is equivalent to:

#(4-3x>=0 and 4(4-3x)=4x+6) or (4-3x<0 and 4(-4+3x)=4x+6)#

that becomes:

#(x<=4/3 and 16-12x=4x+6) or (x>4/3 and -16+12x=4x+6)#

#(x<=4/3 and 16x=10) or (x>4/3 and 8x=22)#

#(x<=4/3 and x=5/8) or (x>4/3 and x=11/4)#

so

#x=5/8 and x=11/4#

Answer 2 · 2016-06-28T14:45:35

#x=5/6 and 11/4#. There is no extraneous solution.

Divide by 4 and make it

#|4-3x|=x+3/2>=0# So, the solution(s) should satisfy #x>=-3/2#.

The given equation is equivalent to the pair

#4-3x=x+1.5#, when # x<=4/3#.and

3x-4=x+1.5, when #x>=4/3.#

Solving both,

x=5/8<4/3 and x=11/4>4/3.

Both are valid..

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