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Method 1

Solve for x $$ax - a^2 = bx - b^2$$

Collect all terms with x on one side of the equation $$ax - bx = a^2 -b^2$$

Factor both sides of the equation $$(a -b)x = (a+b)(a - b)$$

Divide both sides of the equation by the coefficient of $x$ (which is $a-b$)

$$x = a + b$$ (where $a \neq b$ since this would mean dividing by $0$)

Method 2

Solve for $x$ $$ax - a^2= bx - b^2$$

Bring all the terms to one side of the equation $$ax - a^2 -bx + b^2 = 0$$

Rearrange $$ax - bx -(a^2-b^2)=0$$

Factor $$(a - b)x - (a + b)(a - b) = 0$$

$$(a - b)( x - (a + b)) = 0$$

which is a true statement if $$a-b=0$$ $$a = b$$ or $$x-(a+b)=0$$ $$x = a + b$$

My question is I don't understand how this second method is consistent with the first in terms of the restriction on $a$ and $b$.

user 85795
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    And your question is ... ? – Mark Bennet Jun 13 '13 at 22:13
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    The first method assumes $a\ne b$ and solves for $x$ in terms of them. The second method does not assume $a\ne b$, with solutions split into cases - it is simply more general than the first in that it handles more cases. These are consistent because when $a\ne b$ they give the same answer and do not contradict each other (when $a=b$ the first method doesn't apply so does not contradict anything the second method says). – anon Jun 13 '13 at 22:15
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    @skullpatrol Arjang changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Jun 13 '13 at 22:20

3 Answers3

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This is indeed consistent solutions:

  • note: if $a = b$, you have $0 = 0$ (just put $a = b$ and substitute into the original equation).

  • Note, in the first method, you acknowledged that you could not have $a = b$ and validly divide by $(a - b).$ You just forgot to check what happens when $a = b$.

When $a = b$, we have that any $x$ will solve the equation. That is, $a = b \implies x \in \mathbb R$. That doesn't tell you much! $a = b$ is not itself a solution to $x$.

amWhy
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  • Here, though we are sort of treating $x$ as the variable, and $a, b$ as constants. So although we exclude $a - b = 0$ because we cannot divide by zero, we have to consider what happens in terms of $x$ when $a = b$. As it turns out, in both methods, when we consider what happens when $a = b$, we have that there are infinitely many values of $x$ that satisfy the equation when $a = b$. So $a = b$ is not really a solution to $x$ in the second method...it's just that $a = b$ satisfies the equation, regardless of the value of $x$. – amWhy Jun 14 '13 at 01:44
  • To put it briefly, both methods tell us that the only unique solution for $x$ is determined by $x = a+b$. When $a = b$, we know nothing more about $x$ than before we began trying to solve for it: namely, that it is assumed to be some real number. – amWhy Jun 14 '13 at 01:51
  • Yes, you can look at it that way! That would be a good way to summarize what seems to be (but really isn't) discrepancy in results... Alternatively, there are problems when $a = b$, using either method, so we disallow the case $a = b$. – amWhy Jun 14 '13 at 02:00
  • Yes, that's right. By the first method, we do get the true statement $a = b$. Let me explain: In the first method, it is not correct to say that, with respect to the original equation, that $a$ cannot equal $b$. Before deividing by $a - b$, we have to consider what happens when $a$ does equal* $b$? So, we check that out, using the original equation, and if we do that, by substituting b with $a$, we get the true equation $ax - a^2 = ax - a^2 \implies 0 = 0$. So a = b results in one true statement. Then we see what happens when $a\neq b$, and we divide by $a - b$...getting $x = a+b$ – amWhy Jun 14 '13 at 02:39
  • That's what I meant in my answer when I said in your first method, you forgot to check what happens when $a = b$. You are right that we cannot divide the equation by zero. But that doesn't mean a cannot equal b. In the case $a = b$, we simply don't divide...we check that out by substituting one for the other, because in that case, $a = b$. When we've checked that out, then we proceed to solve for all cases $a \neq b$, and since in all those cases, $a\neq b$, we can divide by $(a - b)$. – amWhy Jun 14 '13 at 02:42
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If $a=b$, the left side of the equation $(a-b)(x - (a+b)) = 0$ is always equal to the right side, so $x$ can be anything.

Robert Israel
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In the first method you come to the equation

$$(a-b)x=(a+b)(a-b)$$

In the second method, you come to the equation

$$(a-b)(x-(a+b))=0$$

which is just the first equation rearranged and factored. If we divide the first equation by $(a-b)$ without factoring, we still collect the solution $a=b$, because notice that it does solve the first equation. However, we continue on assuming $a\ne b$ to discover the other solution.

The method is the same with the second equation. We first collect the solution $a=b$, and then we assume $a\ne b$ to see if there are other solutions. By the zero product property, it follows that we must have $x=a+b$.

Jared
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