Solve and check for extraneous solutions
16=3(x-1)-(x-7)

- anonymous

Solve and check for extraneous solutions
16=3(x-1)-(x-7)

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- anonymous

@satellite73

- anonymous

not sure what "extraneous" solutions means in this context
it is a linear equation right? no radials, no denominators

- anonymous

distribute first on the right both the 3 and the minus sign

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## More answers

- anonymous

let me know what you get

- anonymous

Ok, so I got 2x+6=16

- anonymous

think there is a mistake there
the \(2x\) is right, but where did the \(6\) come from?

- anonymous

-1+7

- anonymous

\[3(x-1)-(x-7)=3x-3-x+7\] is a start

- anonymous

oops forgot the distributive law didn't you?

- anonymous

\[3(x-1)=3x-3\times 1\]

- anonymous

Oh okay! So 2x+4?

- anonymous

right
\[2x+4=16\] takes only 2 steps to solve

- anonymous

a) subtract 4
b) divide by 2

- anonymous

2x=10
x=2

- anonymous

???

- anonymous

first off \(16-4\neq 10\) and secondly \(\frac{10}{2}\neq 2\) so both steps are wrong

- anonymous

what is \(16-4\)?

- anonymous

16-4=12

- anonymous

ok so
\[2x+4=16\\
2x=12\] is a the first step

- anonymous

x=6! So sorry, it's late where I am lol so I have a sleepy brain I suppose

- anonymous

second step is \(x=\frac{12}{2}\)

- anonymous

x=6

- anonymous

right

- anonymous

How do we check for extraneous solutions?

- anonymous

there are no extraneous solutions with linear equations, nothing to check

- 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

ok:) Can I ask another? I'll be better haha

- anonymous

sure

- anonymous

x/3=x/2-2
Check for extraneous solutions

- anonymous

is this
\[\frac{x}{3}=\frac{x}{2}-2\] like that ?

- anonymous

yup

- anonymous

you have a choice
you can work with the fractions or you can get rid of them

- anonymous

you pick

- anonymous

get rid of them.

- anonymous

Actually, which ever is simpler

- 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

\[6\times \frac{x}{3}=6\times (\frac{x}{2}-2)\]

- anonymous

I think I did it wrong bc I got 2x/6-3x/6

- anonymous

since \(\frac{6}{2}=3\) and \(\frac{x}{3}=2\) you end up with
\[2x=3x-18\]

- anonymous

you forgot to get rid of the 6 when you cancelled

- anonymous

oh okay:)

- anonymous

\[6\times \frac{x}{3}=2x\] not \(\frac{2x}{6}\)

- anonymous

so now you have
\[2x=3x-18\] which takes 2 steps to solve
you see where the \(-18\) came from?

- anonymous

Yes!

- anonymous

x=18

- anonymous

yes

- anonymous

So it wouldnt be extraneous?

- anonymous

no it would not

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