ab-c=cb and a is not equal to b then b=

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soul88

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I saw this question on a test and i cant seem to figure it out or find the answer. Would appreciate any help. Thanks guys
 
ab-c=cb and a is not equal to b then b=

Have you considered solving for b?

ab - cb = c

(a-c)b = c

It looks more important for \(\displaystyle a\ne c\) rather than \(\displaystyle a\ne b\)
 
soul88 said:
I saw this question on a test and i cant seem to figure it out or find the answer. Would appreciate any help. Thanks guys
Click to expand...

Bring all the terms with 'b' on the left hand side of the equation and all the terms "without b" to the right-hand-side of the equation.

See if you can "factor-out" b.
 
tkhunny said:
ab-c=cb and a is not equal to b then b=

Have you considered solving for b?

ab - cb = c

(a-c)b = c

It looks more important for \(\displaystyle a\ne c\) rather than \(\displaystyle a\ne b\)
Click to expand...

Thanks yes it probably was \(\displaystyle a\ne c\) sorry i couldn't remember exactly how it was on the test. from what you've worked out already i think b = c/(a-c). The trouble i was having is getting to the first step you have. the only way i can think of removing b from right side is to divide by b but then i would end up with ab-c/b = cb/b and all the b's disappear :(
 
Last edited:
Subhotosh Khan said:
Bring all the terms with 'b' on the left hand side of the equation and all the terms "without b" to the right-hand-side of the equation.

See if you can "factor-out" b.
Click to expand...

Ya that is where i'm stumped only way i can think of to remove b from right side is to divide both sides by b but that just cancels out all the b's :(
 
after looking at it for a while I think i get it add c to both sides and subtract bc both sides? but i thought multiplication/division always happened before addition/subtraction.
 
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