solving modulus functions

In the mark scheme, it says"Attempts to solve the correct equation without modulus signs" Why is that so? And when do we need to solve the equation with the modulus function Screenshot 2025-05-17 141640.png

(edited 10 months ago)

Reply 1

mqb2766

Original post

by Noname60

If you sketched the line (16-4x) on the given graph, then it should be fairly easy to see/reason that there must be a single intersection point. As the line lies above the "v" curve and P before the intersection, then the solution of
16-4x = 3|x-2|+5
can be obtained simply by removing the modulus function as x>2 at the intersection point. Id presume they want a proper justification of this, so maybe consider x>=2 and x<=2 seperately and insert the appropriate sign when you remove the mod function. The "note" implies that a justification that the intersection happens for x>2 (as above) would be ok as well, as well as squaring things up and rejecting one of the solutions.

Reply 2

Noname60

Oh I got it now, thank you a lot

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