Question

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

Answer 1

@satellite73

Answer 2

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

Answer 3

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

Answer 4

let me know what you get

Answer 5

Ok, so I got 2x+6=16

Answer 6

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

Answer 7

-1+7

Answer 8

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

Answer 9

oops forgot the distributive law didn't you?

Answer 10

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

Answer 11

Oh okay! So 2x+4?

Answer 12

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

Answer 13

a) subtract 4
b) divide by 2

Answer 14

2x=10
x=2

Answer 15

???

Answer 16

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

Answer 17

what is \(16-4\)?

Answer 18

16-4=12

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

x=6

Answer 23

right

Answer 24

How do we check for extraneous solutions?

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

sure

Answer 29

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

Answer 30

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

Answer 31

yup

Answer 32

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

Answer 33

you pick

Answer 34

get rid of them.

Answer 35

Actually, which ever is simpler

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

you forgot to get rid of the 6 when you cancelled

Answer 41

oh okay:)

Answer 42

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

Answer 43

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

Answer 44

Yes!

Answer 45

x=18

Answer 46

yes

Answer 47

So it wouldnt be extraneous?

Answer 48

no it would not