Solve Basic Math: Find "a" in y = a - ab

  • Thread starter Jasonp914
  • Start date
In summary, to solve for "a" in the equation y = a - ab, we factor out the "a" on the right hand side to get y = a(1-b). Then, we divide both sides by (1-b) to get y/(1-b) = a. It is important to note that when dividing by a variable, we must specify that the result is only true if the divisor is not 0.
  • #1
Jasonp914
7
0
How do I find a? there are two. y = a - ab
 
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  • #2
a - ab is the same as a(1-b)
 
  • #3
simple ans...

y = a - ab

You have two a's on the right hand side (rhs). That's a problem.

So we factor out the "a" on the rhs(right hand side). y=a(1-b).

Now we only have one "a" on the rhs.

a(1-b) means "a" multiplied by (1-b). To solve for "a" we must get rid of (1-b).We do the opposite of multiplying, which is? yes yes yes dividing!

Remember that what we do to one side we must do to the other.

So we get y/(1-b) =a(1-b)/(1-b), (1-b) cancels on the rhs and we finally get

y/(1-b) = a.

:biggrin: :biggrin: :biggrin: :biggrin:
 
Last edited:
  • #4
factor
please post in high school section next time about algerbra
y = a - ab
y= a(1-b) divide (1-b)
y/(1-b) = a
 
  • #5
Don't get snippy, tom. There are some people (English majors?) who didn't take in math in high school (or didn't pay attention when they did) and are paying for it in college. If this was posted by a person who is in college, it belongs in this forum.
 
  • #6
My high school algebra teacher would kill us if we ever divided by a variable. Its simply just bad algebra. While I agree that the only way to solve for a in this case is the way that was illustrated above, you should explicitly write for b not equal to 1.
 
  • #7
I'm 46 and decided to go back to college, they had me take this as a "refresher course" thanks for your help.
 
  • #8
whats wrong with dividing by variables?
 
  • #9
Jasonp914 said:
I'm 46 and decided to go back to college, they had me take this as a "refresher course" thanks for your help.


That's a very brave, usefull and rewarding decision Jason, I applaud you.

I tutor students in their first years of mathematics and physics and I can tell you it ain't easy. Please, do not hesitate if you have any other question. I am willing to help you out if necessary. :wink: :wink:

regards
marlon
 
  • #10
whats wrong with dividing by variables?


You should never divide by a variable (well in 99% of the time) because you don't know what that variable is. You could be very well in fact be dividing by zero. For example x^2=2x. You can't divide both sides by x and get x=2. you have to subtract 2x from both sides and factor out the x and get your solutions as 2 and 0.
 
  • #11
You should never divide by a variable without specifying that the result is only true if the divisor is not 0. As gravenewworld said when asserting that you should not divide by 1-b, " While I agree that the only way to solve for a in this case is the way that was illustrated above, you should explicitly write for b not equal to 1."
 

What is the equation y = a - ab used for?

The equation y = a - ab is used to find the value of "a" when given the values of "y" and "b".

How do I solve y = a - ab?

To solve y = a - ab, you can rearrange the equation to isolate "a" by adding ab to both sides. This will leave you with a = y + ab. Then, you can factor out "a" to get a = (1 + b)y. Finally, divide both sides by (1 + b) to get the value of "a".

What if I don't know the value of "b"?

If you do not know the value of "b", you will not be able to find the exact value of "a". However, you can still solve for a using algebraic manipulation by treating "b" as a variable and following the steps outlined in the previous answer.

Can I use this equation for any value of "y" and "b"?

Yes, this equation can be used for any numerical values of "y" and "b". However, if "b" is equal to -1, the equation becomes y = a + a, which does not have a unique solution. In this case, there are infinitely many possible values for "a".

Can I use this equation for more complex math problems?

Yes, this equation can be used for more complex math problems that involve finding the value of "a" in terms of other variables. However, it is important to make sure that the equation is set up correctly and that any additional variables are accounted for in the equation.

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