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I have a problem asking to find two numbers that have a difference of $4$ and product of $22$. So I set up two equations, $s - t = 4$ and $s\times t = 22$. I then proceeded to solve the first equation for $t$ and got $-t = 4 - s$, I then multiplied both sides by $-1$ and got $t = -4 + s$. The solution in the book I am using solved for $s$ and got $s = t + 4$. Are these two equations equivalent or am I doing something wrong when multiplying both sides by $-1$? The answer to the problem was a negative and a positive number but I got two positives instead.

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What you have shown us is correct, and the two equations are equivalent. The answer should be two positive numbers, so the book must have made some later mistake. – Gerry Myerson Sep 21 '12 at 0:17
Since the product of a negative and a positive is negative, it will not be (positive) 22. The book answer must be wrong. You are on the right track. When you substitute this into the other equation, you will get a quadratic in s. – Ross Millikan Sep 21 '12 at 0:22
I think I see what's going on here. Let $s$ be the larger of the two numbers. Then there are two values for $s$, one positive, one negative. Perhaps it is these two numbers that appear in the book. – Gerry Myerson Sep 21 '12 at 0:37
up vote 1 down vote accepted

You have done well, multiplication by $-1$ is fine. Now, as you want to compare with your book, write down the expression for $s$, in terms of $t$ and 4 i.e., keep the $s$ at the right hand side and move the $-4$ to the left hand side to get $t+4=s$.

Now, as for the solution, surely both the number should be of same sign, otherwise, if they have opposite sign, their product would be negative. It looks like both the numbers can be positive (as well as negative).

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Oho..sorry! Edited. – Tapu Sep 21 '12 at 0:44

$4$ and $8$ have a difference of 4 but $s-t=-4$ and that's not $4$. So I think your first equation should be $|s-t|=4$.

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There is nothing wrong with assuming (as OP implicitly did) that $s$ is the larger of the two numbers. – Gerry Myerson Sep 21 '12 at 0:34
Oh I see. Thanks. – user41728 Sep 21 '12 at 0:55

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