# How can we solve two equations with three unknowns or variab

### How can we solve two equations with three unknowns or variab

how can we solve two equations with three unknowns or variables for ex:

x+y+z=100 &
5x+2y+ 0.25z=100 , then x=?, y=?, z=? where x,y,z are positive integers

Guest

### Re: how can we solve two equations with three unknowns or va

In general with 2 equations and 3 variables you can't find a unique solution, instead you'll usually be faced with infinite solutions which can be written as a function of one variable. However, there is a slight twist in your case that you are told that $x,y,z$ are positive integers, this gives us extra information which can narrow down the number of solutions.

Start as you would normally, by solving for $x,y,z$ in terms of one of the variables (choose your favourite, it shouldn't matter), say $x$. So we solve assuming $x$ is a fixed known constant, so now we have 2 unknowns $y,z$ and 2 equations.
$x+y+z=100 \implies y = 100-x-z$
Substitute this into
$5x+2y+z/4 = 100 \implies 5x+2(100-x-z)+z/4=100 \implies z = (12x+400)/7$
Substitute this back into $y$
$y=100-x-z \implies y=100-x-(12x+400)/7 \implies y=(300-19x)/7$

So $y=(300-19x)/7$ and $z = (12x+400)/7$, remember that the question tells us that $x,y,z>0$, this means that
$y=(300-19x)/7>0 \implies 300-19x>0 \implies 300>19x \implies 300/19=15.789...>x$
Since $x$ is positive and an integer we've narrowed its down to 15 possibilities, $x=1,2,3,...,15$.
We could just try all 15 possibilities and see which ones work. However, if you know a bit of modulo arithmetic you can save yourself a bit of time.

So so far we've used the fact that $x,y$ are positive, if we try to do the same trick with $z>0$, we don't get anything useful. We get $z>0$ as a consequence of $x>0$, so it doesn't help us. The other information we are given is that $x,y,z$ are integers. In particular this means that $y = (300-19x)/7$ must be an integer, or put another way $300-19x$ must be divisible by $7$. Using modula arithmetic this would be represented by $300-19x\equiv 0\bmod 7$, you can look up tutorials on the internet on how to solve this, but it will essentially tell you that if $300-19x$ is divisible by 7 then so is 4 times that, i.e.
$4\times (300-19x) = 1200-76x = 7\times(172-11x)+(x-4)$
If the whole thing is divisible by 7 and $7\times(172-11x)$ is clearly divisible by 7, then it follows that $x-4$ is divisible by 7, i.e. $x$ has remainder 4 upon dividing by 7. This narrows down $x$ to 4, or 11.
We could try the same trick on $z$ being an integer but it turns out it will just tell us again that $x$ has remainder 4 when dividing by 7, so it isn't useful.

So there are two possible solutions $x=4, y=32, z=64$, and $x=11, y=13, z=76$, which you can check both satisfy all the constraints in the question.

Hope this helped,

R. Baber.

Guest

### Re: How can we solve two equations with three unknowns or va

thank you Mr, Baber for your help

a customer buys some apples, mangos and oranges which total comes exactly 100 fruits

the cost of one apple is 5 Rs.
the cost of one mango is 2 Rs.
and four oranges cost 1 Rupee.

the total amount he pays also equal to 100 Rs.

then how many apples, mango and oranges did he buy?

how these kinds of problems solved mathematically?

Guest