How to find all the positive integral solutions of $5x+7y=100$? How to find the number of all the positive integral solutions and the solutions itselves of $$5x+7y=100?$$
Please help me!!
 A: Assume we have $5x+7y=100$. Let us solve for the variable with the smallest coefficient $x=20-\frac{7y}{5}$. This means that $\frac{7y}{5}$ is integer. Therefore $y=5y_2$ for some integer $y_2$. 
Plugging this into the original equation we get $x=20-7y_2$. You see now, that for every integer you put in $y_2$ you get an integer value for $x$, and an integer value for $y=5y_2$ that satisfy the equation $5x+7y=100$.
Since we want positive solutions you can put $y_2>0$ such that $20-7y_2>0$, i.e. $20/7>y_2$. So, $0<y_2\leq 2$. 
We have then two possibilities: $y_2=1$ or $y_2=2$, for which we get $y=5$, or $y=10$ respectively. The corresponding values for $x$ are $x=13$, or $6$.
A: Given $$5x+7y=100$$ also $x\ge0$ and $y\ge0$ie you get the inequalities $$5x\le100$$ or $$x\le20$$ similarly $$y\le14.28$$
Now $$x=\frac{100-7y}{5}$$ which means $5$ divides $7y$. The only values satisfying for $y$ are $$y=0,5,10$$ and is then bounded by the inequality.
A: positive integral means that positive and integer right?because $5$ and $7$ are coprime,that means that $100$ is represented by sum  of two number,which one  of them is divisible by $5$ and $7$,one of possible solution is $30+70$,which is equal  to exactly $100$, or  $x=6$ and $y=10$,could you found others? 
EDITED:
second is just $35+65$
there $35$ is divisible by $5$ and $7$
that means that $y=5$ and $x=13$
generally  $x=(100-7*y)/5$
how much positive integer of $y$ is so that $(100-7*y)$ is divisble by $5$ and leaves $x$ positive?
clearly $y$ is positive number,which is multiply of $5$
A: Hint $\ $ Since $\,5x+7y\,$ is linear in $\,x,y,\,$ the general solution of $\,5x+7y=100\,$ is the sum of any particular solution, e.g. $\,(x,y)=(20,0),\,$ plus the general solution solution of the associated homogeneous equation $\,5x+7y = 0,\,$ which is $\,(x,y)=(-7n,5n).\,$ Summing them gives the general solution  $\,(x,y) = (20,0)+(-7n,5n) = (20-7n,5n).\,$ Now $\,5n > 0\!\iff\! n\ge 1,\,$ and $20-7n>0\!\iff\! n\le2.$
A: $5x+7y=100~\implies 5|y$
$y$ is limited to 14 on the upper side. So, we are left with only $y=5,10$
When,$y=5$,we can have a solution,i.e.,$x=13$
Also,$y=10$,we can have a soluton,i.e.,$x=6$  
A: $5x \gt0$  so $x=1,2,3,4,5,\ldots $
careful about $5x$  and 100 both are 5q ---> so 7y must be 5K too
then put $y=0,5,10$ (not more because $7y$ must be less or equal to $100$)
so 
check them to find $x $
A: Since $5 \mid 100$, we know that $5$ must divide $5x +7y$ and that therefore $5$ must divide $7y$. Since $5$ and $7$ are coprime, only multiples of $5$ as $y$ will suffice, and since $15 \cdot 7>100$ and both $x,y$ are integral and positive, only $y=5,10$ are solutions. 
We have now shown that $(13,5)$ and $(6,10)$ are the only solutions.
Edit: $(20,0)$ can also be a solution if you consider  $0$ a positive number
A: As $(5,7)=1$ we would definitely have some $x,y\in \mathbb{Z}$ such that $5x+7y=1$
we see that $5(3)+7(-2)=1$ i.e., $5(300)+7(-200)=100$
Now if $ax_1+by_1 = c$ is any solution, then all solutions are of the form
$$x = x_1 - r\frac{b}{\gcd(a,b)},\qquad y = y_1 + r\frac{a}{\gcd(a,b)}$$
we have $(5,7)=1$ so any solution would be of the form 
$x=300-7r$ and $y=-200+5r$
We want solutions to be positive which means that we want $7r<300$ and $5r>200$ 
i.e., $r<43$ and $r>40$ i.e., $r\in \{41,42\}$
For $r=41$ we would have : 
$x=300-287=13; y=-200+215=5$ i.e., $(x,y)=(13,5)$
For $r=42$ we would have : 
$x=300-294=6; y=-200+220=10$ i.e., $(x,y)=(6,10)$
So, only positive solutions are $(13,5)$ and $(6,10)$. 
If you consider $0$ to be positive then $(20,0)$ would also be a solution you are looking for 
A: Thank you for pointing out my mistake, postmortem. 
Through diophantine equation, I get 
$$x = 300 -7r,$$ 
for all $r-$ integers, 
$$y = -200 +5r $$
To solve the inequalities above. So, to get the positive solution, for the above inequalities, 
we need to set certain boundaries such that, 
$r\ge40$ , and $r<42$, 
sorry for my mistake, due to I am having fever. Thank you so much for pointing out my mistake. 
Reference：
How to find solutions of linear Diophantine ax + by = c?
