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conceptual algebra question -- instructors!?!

Problem Solving — algebra and arithmetic
Expert replies
by san2009 » Thu Jul 29, 2010 9:25 am
if you have
(x-1)(x-2)=(x-2)
then CAN YOU cancel out the (x-2) on either side and get x-1 = 1 or x=2 as the solution
OR ....
is the correct way of solving this
(x-1)(x-2)=(x-2)
x^2-2x-x+2=x-2
which gives you the quadratic eqt
x^2-4x+4=0
and when you factor that, you get
x=2 as the only solution

Expedited feedback appreciated.
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Source: — Quantitative Reasoning |

by Tani » Thu Jul 29, 2010 1:52 pm
Either is acceptable - both give you x=2 as the only solution.


The first way you get x-1 = 1, or x-2 = 0. Add one to each side of the first equation and you get x = 2.
Add 2 to each side of the second equation and you also get x = 2.
Tani Wolff
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by Ian Stewart » Fri Jul 30, 2010 9:50 pm
san2009 wrote:if you have
(x-1)(x-2)=(x-2)
then CAN YOU cancel out the (x-2) on either side and get x-1 = 1 or x=2 as the solution
OR ....
is the correct way of solving this
(x-1)(x-2)=(x-2)
x^2-2x-x+2=x-2
which gives you the quadratic eqt
x^2-4x+4=0
and when you factor that, you get
x=2 as the only solution

Expedited feedback appreciated.
No, you can't simply cancel the x-2 from both sides, since x-2 might be equal to zero, and we can never divide by zero. For example, if you have the equation:

3x = x

you can't simply cancel the x to find that 3 = 1 (that's clearly nonsense). Here the only solution is x=0.

So, in your example above, if you simply cancel the x-2, you risk losing one of the solutions to the equation. Now, it turns out not to matter in your example, but that's pure coincidence. If instead you had the equation

(x-2)(x) = x-2

if you were to simply cancel the x-2 terms, you'd only find one solution: x=1. You'd miss the solution x=2. To be sure you don't miss solutions, you can either expand and factor, or you can cancel the x-2 as long as you consider the possibility that x-2=0 gives you a solution to your equation.
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by san2009 » Sat Jul 31, 2010 12:34 am
excellent. thanks Ian!
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by Tani » Sat Jul 31, 2010 7:13 am
If by canceling (x-2) you mean divide by (x-2), you get X-1 = 1, not x-1 = 0 . That gives you x=2, not x=1. x=1 is not a solution. Plug x=1 into the original and you get 0 = -1
Tani Wolff
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by Ian Stewart » Sat Jul 31, 2010 9:32 pm
Tani Wolff - Kaplan wrote:If by canceling (x-2) you mean divide by (x-2), you get X-1 = 1, not x-1 = 0 . That gives you x=2, not x=1. x=1 is not a solution. Plug x=1 into the original and you get 0 = -1
You may have misunderstood my post. When I discussed the solution x=1, I was discussing the equation I presented in my post, not the equation in the original post.

The equation in the original post produces one of those rare quadratics which has only one solution, x=2, which makes it a very poor equation to look at if you want to understand why it is not acceptable to simply divide both sides of an equation by x-2 -- you happen to get the right answer if you do in this one case, but that's just luck. In general, you will not get all of the solutions to an equation if you divide by unknowns on both sides, unless you consider that you might be dividing by zero. The simplest example is the following:

x^2 = x

Here, we can't just divide by x to find that x=1. It's also possible that x=0.
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by shashank.ism » Wed Aug 04, 2010 12:16 am
See for these type of problems before cancelling we go on considering two cases...
case1) x=2 ...so its a clear solution...
case2) x=/=2, so in this case we can cancel x-2 on both sides ,...so x-1=1 --> x=2.

hence the solution is x=2


if the question might have been (x-4)(x-2)= (x-2)
Now again we go on by considering two cases...
case1) x=2 ...so its a clear solution...
case2) x=/=2, so in this case we can cancel x-2 on both sides ...so x-4=1 --> x=5.
so solution in this case is x=5,2
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