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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?

  • 3
    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.2012-11-08
  • 0
    There are still infinitely many solutions, even if you require them to be positive integers.2012-11-08
  • 0
    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.2012-11-08
  • 0
    Can any one show how you can get 15 and 9 as the answers?2012-11-08

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