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well, sometimes when you solve an equation and end up with an "x" value, the x-value doesn't make the equation true when you plug it back in, because it isn't in the domain of the original function in short: an extraneous solution is like a "fake" solution that you find that doesn't make the equation true when you plug it back in
so, looking at our problem, do you know how to solve for x?
can you subtract 1/(x+4)?
well, technically yes, but in this case we want to try a different method this equation doesn't have any real (non-extraneous solutions) but we want to find the "fake" extraneous one
okay then idk
so, we multiply everything by (x+4) to get rid of the denominator
that gives us 1 + (1/2)(x+4) = 1 can you try solving for x now?
First can you subtract the one?
yes, just treat it like a normal equation
how do you multiply 1/2 by x+4
you don't have to after you subtract 1 from both sides, we have (1/2)(x+4) = 0 now divide both sides by (1/2), what do you get?
x+4=0 lol sorry i knew that i don't know why i forgot You get -4 right?
so its B?
nope, not B, hang with me for a sec ;) x = -4 is called an "extraneous solution" although we got x = -4 as an answer, when we plug x = -4 back into the original equation, we can see that the denominator of the first fraction becomes 0, x = -4 isn't a "real" solution
can u help me with a few more?
sure, post a new q and tag me