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Can you please help me identify where I went wrong, for the following question, when I plug in my answers using substitution method I get the correct answer, but I found out that it could also be $15$ and $9$ but how do you get that.

$$14x - 5y + 14 = 179  \\ -14x + 5y + 5 = -160.$$

I rearranged the first equation,
$14x - 5y + 14 = 179\\  14x = 5y -14 +179\\ x = \frac{5}{14}y + \frac{165}{14}.$

Then using the second equation, I solve for $y$.
$-14x+5y+5=-160\\ 5y+5=160+14x\\ 5y=-165+14x\\ y=-33+\frac{14}{5}x$

Sub the $x$ value which we got before
$y=-33+14/5(165/14+5y/14) \Rightarrow\\ y=1.$

As $y=1$, sub this into the second equation,
$-14x + 5(1)+5 =-160\\ -14x= -170\\ x= 12.142857$
to 5 decimal places $= \ 12.14286$

$14(12.14286)-5(1)+14 = 179.00004\\ -14(12.14286)+5(1)+5 = -160.00004$

Where did I go wrong?

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I don't know how you got $y=1$, but your problem is those two equations are actually the same. This means that you will have infinite solutions. – Javier Nov 8 '12 at 11:34
There are still infinitely many solutions, even if you require them to be positive integers. – Gerry Myerson Nov 8 '12 at 11:49
I would also recommend you write your $x$ as $\frac{85}{7}$ rather than using an abbreviated decimal expression. Then you don't get those rounding errors when you make the substitution. – Matthew Pressland Nov 8 '12 at 11:50
Can any one show how you can get 15 and 9 as the answers? – Jay Nov 8 '12 at 11:55

First off, you made a mistake when finding the value of $y$. $y=-33+\frac{14}{5}(\frac{165}{14}+\frac{5}{14}y)$ is correct, but if you expand everything you get $y = -33 + 33 +y$. This reduced to $0=0$ which, while correct, is certainly not very useful.

The problem here is that you have these two equations:

$$14x - 5y + 14 = 179 \\ -14x + 5y + 5 = -160 $$

Rearrange to separate unkowns and numbers:

$$14x-5y = 165 \\ -14x+57 = -165$$

Multiply the bottom equation by $-1$:

$$ 14x-5y = 165 \\ 14x-5y = 165 $$

See? What looked like two equations is actually only one written differently. From this equation you could, for example, solve for $y$:

$$y= \frac{14}{5}x-33$$

This tells you that you have infinite solutions; just pick a value of $x$ and you'll get a suitable value of $y$ that fulfils the equation. If you take $x=15$, you get $y = \frac{14}{5}\times 15 - 33 = 42-33 = 9$.

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But you see the answer I obtained is that not correct mathematically. – Jay Nov 8 '12 at 12:30
Only in the sense that you made an error that made it look like $y$ must always be $1$, which isn't true. But if you take $y=1$, then Javier's formula gives you a corresponding value for $x$, in this case $\frac{85}{7}$, so that you get a solution to the equations. Exactly as you can pick any $x$ and get a suitable $y$ to make a solution, as Javier says, you can pick any $y$ and get a suitable $x$. In short, your solution is mathematically correct, despite the error in the way you obtained it. – Matthew Pressland Nov 8 '12 at 12:56

In the comments, OP asks, "Can any one show how you can get 15 and 9 as the answers?"

First, as Javier points out, you show that your two equations are the same, so you really just have to find a solution to a single equation, $$14x-5y=165.$$ Second, you make an executive decision that you are only interested in whole number answers (or maybe the person who gave you the problem tells you that she is only interested in whole number answers). In that case, since $165$ is a multiple of $5$, and $5y$ is a multiple of $5$, $14x$ must also be a multiple of $5$; and since $14$ is not a multiple of $5$, which is a prime number, $x$ must be a multiple of $5$. That is, $x=5Q$ for some whole number $Q$. So our equation becomes, $$(14)(5Q)-5y=165,$$ and we can divide through by $5$ to get $$14Q-y=33,$$ which we can rewrite as $$y=14Q-33.$$ Now you can let $Q$ be any number you like, and this formula will tell you what $y$ has to be (and $x=5Q$ will tell you what $x$ has to be). Since you decided that you want whole numbers, you had better pick $Q$ to be a whole number, and you had better pick it large enough to make $14Q-33$ a whole number. A little experimentation (or dividing $33$ by $14$) will convince you that the smallest number that works for $Q$ is $Q=3$. Then $x=5Q=15$, and $y=14Q-33=9$, and that's how you can get 15 and 9 as answers.

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Why do you decide that you only want integers? I don't see that in the question. – Javier Nov 9 '12 at 2:06
Javier, it's not in the question, but it is in OP's comment. OP specifically asks "how you can get 15 and 9 as the answers." If you know a way to get 15 and 9 as the answers, without somewhere along the way deciding that you only want integers (and positive integers, at that), I'm sure OP would like to see it. – Gerry Myerson Nov 9 '12 at 3:28
I'm sorry if there's something I'm not getting, but couldn't you just set $x=15$, for example, and solve for $y$?. It's not like $(15,9)$ is the only solution, even if you just consider integers, so I don't see how can it be the answer. – Javier Nov 9 '12 at 9:50
@Javier, I think we're both trying to figure out what OP wants, and OP isn't saying. Until OP chimes in, our discussion is moot. – Gerry Myerson Nov 9 '12 at 11:27

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