Algebraic equations

[VeeN22]

Guys help me out
The question says I must solve for 'x'
px - kx = k - p

How do I start?

 

[LCKurtz]

Factor x out of the left side.

 

[Prof]

Find a common factor on the left side of teh equation and factor it out.

 

[Steven G]

In px you have p x's. In kx you have k x's.
So how many x's are there in px-kx?

 

[VeeN22]

Factor x out of the left side.

Fine I have
x(p - k) = k - p
x(p - k) / x = k - p / x

Therefore
p - k = k - p / x

??

 

[LCKurtz]

Fine I have
x(p - k) = k - p
x(p - k) / x = k - p / x

Bad next step. You want the x alone on the left. Divide by the other factor, not the x.

 

[VeeN22]

Bad next step. You want the x alone on the left. Divide by the other factor, not the x.

Thank you ? I think I can take it from there.

 

[pka]

Guys help me out
The question says I must solve for 'x'
\(px - kx = k - p\)

Can't you see by inspection that \(x=-1\) is the solution?

 

[VeeN22]

Can't you see by inspection that \(x=-1\) is the solution?

No. How do you do it?

 

[pka]

No. How do you do it?

I just look at the question and saw that \(x=-1\) makes the equality true.

 

[Steven G]

What do you multiply p-k by to get k-p?

 

[VeeN22]

What do you multiply p-k by to get k-p?

You multiply by -1

 

[Steven G]

You multiply by -1

So if you have x(p-k) = k-p then x= -1. After all x is being multiplied by p-k to get k-p. And you said that only -1 times p-k equals k-p. So x=-1. OK??

 

[VeeN22]

So if you have x(p-k) = k-p then x= -1. After all x is being multiplied by p-k to get k-p. And you said that only -1 times p-k equals k-p. So x=-1. OK??

Yes, thank you.