anonymous
  • anonymous
Solve and check for extraneous solutions 16=3(x-1)-(x-7)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
not sure what "extraneous" solutions means in this context it is a linear equation right? no radials, no denominators
anonymous
  • anonymous
distribute first on the right both the 3 and the minus sign

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anonymous
  • anonymous
let me know what you get
anonymous
  • anonymous
Ok, so I got 2x+6=16
anonymous
  • anonymous
think there is a mistake there the \(2x\) is right, but where did the \(6\) come from?
anonymous
  • anonymous
-1+7
anonymous
  • anonymous
\[3(x-1)-(x-7)=3x-3-x+7\] is a start
anonymous
  • anonymous
oops forgot the distributive law didn't you?
anonymous
  • anonymous
\[3(x-1)=3x-3\times 1\]
anonymous
  • anonymous
Oh okay! So 2x+4?
anonymous
  • anonymous
right \[2x+4=16\] takes only 2 steps to solve
anonymous
  • anonymous
a) subtract 4 b) divide by 2
anonymous
  • anonymous
2x=10 x=2
anonymous
  • anonymous
???
anonymous
  • anonymous
first off \(16-4\neq 10\) and secondly \(\frac{10}{2}\neq 2\) so both steps are wrong
anonymous
  • anonymous
what is \(16-4\)?
anonymous
  • anonymous
16-4=12
anonymous
  • anonymous
ok so \[2x+4=16\\ 2x=12\] is a the first step
anonymous
  • anonymous
x=6! So sorry, it's late where I am lol so I have a sleepy brain I suppose
anonymous
  • anonymous
second step is \(x=\frac{12}{2}\)
anonymous
  • anonymous
x=6
anonymous
  • anonymous
right
anonymous
  • anonymous
How do we check for extraneous solutions?
anonymous
  • anonymous
there are no extraneous solutions with linear equations, nothing to check
anonymous
  • anonymous
you can replace x by 6 in the original equation and see that it works if you like but it is right, nothing is extraneous here
anonymous
  • anonymous
ok:) Can I ask another? I'll be better haha
anonymous
  • anonymous
sure
anonymous
  • anonymous
x/3=x/2-2 Check for extraneous solutions
anonymous
  • anonymous
is this \[\frac{x}{3}=\frac{x}{2}-2\] like that ?
anonymous
  • anonymous
yup
anonymous
  • anonymous
you have a choice you can work with the fractions or you can get rid of them
anonymous
  • anonymous
you pick
anonymous
  • anonymous
get rid of them.
anonymous
  • anonymous
Actually, which ever is simpler
anonymous
  • anonymous
then since the least common multiple of 2 and 3 is 6, if you multiply both sides by 6 they will be gone most people i think like getting rid of fractions
anonymous
  • anonymous
\[6\times \frac{x}{3}=6\times (\frac{x}{2}-2)\]
anonymous
  • anonymous
I think I did it wrong bc I got 2x/6-3x/6
anonymous
  • anonymous
since \(\frac{6}{2}=3\) and \(\frac{x}{3}=2\) you end up with \[2x=3x-18\]
anonymous
  • anonymous
you forgot to get rid of the 6 when you cancelled
anonymous
  • anonymous
oh okay:)
anonymous
  • anonymous
\[6\times \frac{x}{3}=2x\] not \(\frac{2x}{6}\)
anonymous
  • anonymous
so now you have \[2x=3x-18\] which takes 2 steps to solve you see where the \(-18\) came from?
anonymous
  • anonymous
Yes!
anonymous
  • anonymous
x=18
anonymous
  • anonymous
yes
anonymous
  • anonymous
So it wouldnt be extraneous?
anonymous
  • anonymous
no it would not

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