How do you solve $5x – 4 = |2x + 1|$ and find any extraneous solutions?

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Answer 1 · 2017-10-25T18:41:32

$x=5/3$

The solutions for $5x – 4 = |2x + 1|$

are contained in the set of solutions for $(5x-4)^2-(2x+1)^2 =0$

but

$(5x-4)^2-(2x+1)^2=(3 x-5) (7 x-3)=0$ so

$x={3/7, 5/3}$ and the feasible solution is $x=5/3$ because it verifies the original relationship.

How do you solve 5x – 4 = |2x + 1|5x – 4 = |2x + 1| and find any extraneous solutions?