A student returning from Europe changes her Polish z lotys and Czech crowns into Australian money. If she receives $7.90 and has received 44c for each Polish z lotys and 7c for each Czech crown, what amounts of each type of currency did she exchange, given that she started with at least 10 of each?
The equation: $44x + 7y = 790.$
$d = \gcd(44,7)=1.$
I found the solution of $44x + 7y = 1,$ which is $x=1, y=-6.$
By multiplying this solution by $790$, I found one of the solutions for $44x+7y=790$, i.e. $x = 790$ and $y=-4740.$
The general solutions are as follows: $x = 790 + 7n$ and $y = -4740 - 44n$. solving for $n$, I got two values for $n$:
$n = -112$ and $n = -107$
and I solved for $x$ and $y$ and I got $x=34$ and $y=12.$ the problem is I don't know where to use the at least $10$ part of the question. and when I plug in the $x$ and $y$ values to the main equation the answer $790$ doesn't come so I figured it was wrong. Can someone help me with this? Thanks
Probably you just have to fix your first error and the rest will be clear. A solution to $44x+7y = 1$ is $x=4, y=-25$. So the general solution to $44x+7y = 790$ is $x=4\cdot 790 +7n$, $y = -25\cdot 790 -44n.$ Set these last two expressions greater than or equal to $10$ and there is only one value of $n$ that works.