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Mathematics 16 Online
OpenStudy (anonymous):

solve bx-cx=7 for x

OpenStudy (solomonzelman):

Do you know anything like the folowing rule? \(\large\color{black}{ \displaystyle ab+zb=(a+z)b }\)

OpenStudy (solomonzelman):

Don't be afraid to say no....

OpenStudy (anonymous):

I really dont know

OpenStudy (solomonzelman):

it is fine.

OpenStudy (solomonzelman):

Ok, would you agree with me if I said that: \(7x-4x=3x\) ?

OpenStudy (anonymous):

no hold on

OpenStudy (anonymous):

Here is the full question, with options

OpenStudy (solomonzelman):

I don't really need that to do it..... I am just trying to get you to understand how to work ouy the solution:) So, back to what I asked, if you will. In general: \(7x-4x=3x\) Do you agree with me or not?

OpenStudy (anonymous):

I do

OpenStudy (solomonzelman):

Ok, and same way: \(10x-2x=8x\) \(9x-3x=6x\) and so forth.

OpenStudy (solomonzelman):

Now, when you do \(bx-cx\) you are going to have a "b-c" number of x's. Right?

OpenStudy (anonymous):

Yes

OpenStudy (solomonzelman):

So, you can re-write this: \(bx-cx=7\) as: \((b-c)\times x =7\)

OpenStudy (solomonzelman):

good with the step that I did just now, or you have questions about it?

OpenStudy (anonymous):

Good with the steps

OpenStudy (solomonzelman):

Ok, so we have so far: \((b-c)\times x=7\) ---------------------- If I had \(4x=8\), and I wanted to solve for x, THEN, I would divide both sides by 4. \(4x\div4=8\div 4{~~~~~}\Rightarrow{~~~~~} x=2\)

OpenStudy (solomonzelman):

So, what whould I do when instead of 4 (as I showed in the example), I have a \("b-c"\) (as we have in our problem now).

OpenStudy (anonymous):

Alright I sorta got it

OpenStudy (solomonzelman):

Ok, s when we have: \((b-c)\times x=7\) we are going to divide both sides by: \((b-c)\)

OpenStudy (solomonzelman):

Can you preform this division for me please?

OpenStudy (solomonzelman):

perform*

OpenStudy (anonymous):

x/(b-c) ?

OpenStudy (solomonzelman):

x=7/(b-c)

OpenStudy (anonymous):

Thank You :)

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