Word problem, system of equations - Three Variables

Question

Trying to solve this problem.

I've tried a few ways with no luck including substituting, setting some of the equations to zero. Really just trying to understand the material.

The problem reads like this system of equations - am I way off?

x + y + z = 50
20x + 50y = 0.5
30y + 80z = 0.6

Tim wants to buy a used printer. There are three different types to choose from. If all three are used, the time it takes to finish 50 minutes. If printer X operates for 20 minutes and copier Y operates for 50 minutes, one-half of the job is completed. If printer Y operates for 30 minutes and printer Z operates for 80 minutes, three-fifths of the job is completed. Which is the fastest printer on its own? How long does it take this printer to complete the entire job working alone?

Answer 1

Your system of equations should be $$ 50x + 50y + 50z = 1, \\ 20x + 50y =0.5, \\ 30y + 80z =0.6. \\ $$ Here $x$ is the fraction of the "whole job" completed by device $X$ in a minute; similarly for $y$ and $z$.

One way to solve this system is to express both $x$ and $z$ in terms of $y$ from the 2nd and 3rd equation: $$ x = {1\over20}(0.5-50y), \\ z = {1\over80}(0.6-30y); $$ substitute these expressions into the 1st equation (which will now have $y$ only); solve for $y$, then compute $x$ and $z$.

The solution is $$ x={1\over120}, \qquad y={1\over150}, \qquad z={1\over200}. $$

That is, device $X$ is the fastest, and $Z$ is the slowest: ($x>y>z$).

Answer 2

Let $x$ equal the time it takes for printer X to do the job, and $y$ equal etc.

The first thing to notice is that

$x + y + z$ does !!!!!NOT!!!! equal $50$. If all three are running at the same time then then the result must be less than any printer running by itself so $50 \le \min(x,y,z) < x + y + z$.

Instead let $a=$ the number of sheets of paper printer X can print in a minute. $b=$ the number of sheets of paper printer Y can print in a minute. $c=$ the number of sheets of paper printer Z can print in a minute.

And let $N = $the number of sheets of paper in the job.

So $50a + 50b + 50c= N$ is the proper equation.

The next equation is $20a + 50b = \frac 12 N$

And the last equation is $30b+ 80 c = \frac 35 N$.

Can you take it from there.

Answer 3

Very simple (despite two answers given I hope this brings perfect clarity on how to solve your type of problem) . Printer $1$ prints at speed $x$ and the other two at $y$ and $z$ respectively. It means at its speed $x$ printer X finishes $1/x$ part of the job per minute. So, printer X will finish the whole job in $x$ minutes.

If all three are used the job gets done in $50$ minutes:

$50\cdot(1/x+1/y+1/z)=1$

If printer X and Y are used, then we have

$20\cdot(1/x)+ 50\cdot(1/y)=1/2$

And third variation:

$30\cdot(1/y)+ 80\cdot(1/z)=3/5$

Now, let’s get rid of fractions by denoting, say, $1/x=u$, $1/y=v$, $1/z=w$, and we’ll have a system: $50(u+v+w)=1$

$20u+ 50v=1/2$

$30v+ 80w=3/5$

Now you can easily solve it. For example, find $v$ and $w$ in terms of $u$ from the second and third equation. Then plug it into the first equation and find $u$. Then you find speeds $x$, $y$, and $z$. The printer X should be the fastest! Careful! Z should be the slowest. Use common reasoning too. Don’t confuse speed with how much time a printer needs to get the job done. The more time a printer requires to get the job done, the slower it is!