Question

is this right?

Answer 1

(6x)/(3x) = x-5
(6x)/(3x) x 3 = x - 5 x 3
6x = 3x -15
6x - 3x = -15
3x = -15
x =?

Answer 2

not sure if i have worked it out right

Answer 3

(6x)/(3x) x 3 = x - 5 x 3
6x = 3x -15 <<== On the left is should be just 6. (the x's cancel out)

Answer 4

oh right ok

Answer 5

3 also divides into 6. So 2 + 2 - 5.
Add 5 to both sides.
x =7

Answer 6

\[\frac{6x}{3x} = (x-5)\]\[\implies 6x = 3x(x-5), x \ne 0\]\[\implies6x = 3x^2 - 15x, x\ne 0\]
etc.

Answer 7

You get a quadratic, but one of the answers is erroneous because x cannot be 0 (you were dividing by it originally).

Answer 8

hmm, im confused

Answer 9

\(\frac{6x}{3x} = (x-5)\Rightarrow 2 = x-5 \Rightarrow x = 7\)

Answer 10

how did you get from 2 = x - 5 > x = 7

Answer 11

2+5 = x?

Answer 12

\[6x = 3x^2 - 15x \implies 3x^2 - 21x= 0\implies 3x(x-7) = 0\]
So x must be 0 or 7 and it can't be 0 cause you divided by it.

Answer 13

how did you get \[3\times^{2}\]

Answer 14

When you multiply the 3x by the right side of the equation..
\[3x(x-5) = 3x^2 - 15x\]

Answer 15

totally confused never done anything like this before

Answer 16

Ok lets go again step by step. You had the right thought initially.

Answer 17

\[\frac{6x}{3x} = (x-5)\]
So we multiply both sides by 3x.\[\frac{6x}{3x} (3x)= 3x(x-5)\]
Then we expand the terms on the right and simplify the fraction on the left\[6x = 3x^2 - 15x\]

Then we subtract 6x from both sides.\[0 = 3x^2 - 21x\]
We can factor a 3 and an x from each term on the right and we get\[3x(x-7) = 0\]

Answer 18

Now from this we can see that if the whole product equals 0, either x must equal 0 or x must equal 7. Since x was in the denominator originally, x cannot be 0. So it must be 7.

Answer 19

so the last part 3x(x−7)=0 is that needed to be worked out or is it the answer?

Answer 20

It gives us the 2 possible answers.

The only time a product (two numbers multiplied together) is 0 is when one of those numbers is also 0 right?

Answer 21

yep

Answer 22

So in this case you have 3x times (x-7) equals 0.
That means either

3x = 0 OR x-7 = 0

Answer 23

So that gives you

x = 0 OR x = 7

And we know x cannot be 0 because it was in the denominator originally.

Answer 24

Ack! After all that work I just realized there is a much easier way to solve this ;p

Answer 25

\[{6x \over 3x} = x-5\]
Factor a 2 from the numerator on the left..
\[2(\frac{3x}{3x}) = x - 5\]

3x/3x = 1 so that means:
\[2 = x - 5 \implies x = 7.\]

Answer 26

Which I think is the solution the other person here was hinting at.

Answer 27

what is the "factor a 2 from the numerator" mean? eg. how did you do it?

Answer 28

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

Answer 29

so when 3x / 3x = 1

Answer 30

you did ? 2 x 1 = x-5?

Answer 31

right.

Answer 32

and then 2 = x-5 how did that get to 7?

Answer 33

2+5 = x?

Answer 34

add 5 to both sides, yes.

Answer 35

and how did you known to +5?

Answer 36

Because you want to get the x by itself.

Answer 37

know*

Answer 38

So you want to make the -5 into a 0, you add 5 to it.

Answer 39

ah right ok

Answer 40

Nice job. =)