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We have absolute value equations with unknown coefficients: $$|x + a| = b$$ and we know the solutions: $$x = 11 \text{ and } x = 5$$

We need to find $a$ and $b$. From $$11 + a = b \\ 5 + a = -b$$ we get $a = -8$ and $b = 3$.

But we can try another way: $$11 + a = -b \\ 5 + a = b$$ and get $a =8$ and $b = -3$, which is not correct, apparently we need take absolute value from $b$.

How can I formulate this rule and explain it to another person?

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    $\begingroup$ Shouldn't the first solution be $a = -8$? $\endgroup$ – shardulc May 26 '16 at 3:42
  • $\begingroup$ @shardulc, correct $\endgroup$ – Alexan May 26 '16 at 11:32
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We want to solve $|x + a| = b$ where we know that the solutions for $x$ are 11 and 5. Since it's an absolute value, $b$ must be positive and therefore $b > -b$. We might try to solve $$11 + a = -b \qquad 5 + a = b$$ But $11 + a > 5 + a$ and $-b$ cannot be greater than $b$. So we are left with $$11 + a = b \qquad 5 + a = -b$$ which can be solved using simultaneous equations to get $a = -8$ and $b = 3$.

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Plug $x=11$ into the original equation and square both sides. Do the same with $x=5$. We obtain $$(11+a)^2=b^2=(5+a)^2.\tag1$$ Expanding (1) and solving for $a$ we get $$121+22a+a^2=25+10a+a^2\ \Longleftrightarrow\ a=-8$$ Plugging this value of $a$ back into the original equation, this gives $$b=\vert 11+a\vert=\vert 5+a\vert=3.$$

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