Use exact arithmetic to solve the following system?

[Math10]

Homework Statement

Use exact arithmetic to solve the following system:

10^-3x-y=1,
x+y=0.

Homework Equations

I know that 10^-3=0.001.

The Attempt at a Solution

Here's the work:

y=-x
10^-3x+x=1
x(10^-3+1)=1
x=1/(0.001+1)=1/1.001
y=-1/1.001 since y=-x

The answer for this problem is (1/1.001, -1/1.001) but this is a Linear Algebra problem. The work that I've shown above are Algebra 1 skills. I must use exact arithmetic to solve the system. How do I do so?

 

[Ray Vickson]

Homework Statement

Use exact arithmetic to solve the following system:

10^-3x-y=1,
x+y=0.

Homework Equations

I know that 10^-3=0.001.

The Attempt at a Solution

Here's the work:

y=-x
10^-3x+x=1
x(10^-3+1)=1
x=1/(0.001+1)=1/1.001
y=-1/1.001 since y=-x

The answer for this problem is (1/1.001, -1/1.001) but this is a Linear Algebra problem. The work that I've shown above are Algebra 1 skills. I must use exact arithmetic to solve the system. How do I do so?

Do NOT use decimal; use exact fractions instead. Your original equations are
[tex] \frac{x}{1000} - y = 1 \\
x + y = 0[/tex]
So, of course, you have
[tex] \frac{x}{1000} + x = 1 \: \Longrightarrow \left( \frac{1}{1000} + 1 \right) x = 1[/tex]
Surely you can finish this!

 

[HallsofIvy]

Frankly, I don't see any problem with using 1.001. That's every bit exact as [itex]\frac{1001}{1000}[/itex].

If you had written something like 0.3333 for 1/3, then you would not be using "exact arithmetic".

 

[Ray Vickson]

Frankly, I don't see any problem with using 1.001. That's every bit exact as [itex]\frac{1001}{1000}[/itex].

If you had written something like 0.3333 for 1/3, then you would not be using "exact arithmetic".

I was trying to offer advice that would apply to any, general system, not just to the very special (and very simple) example case illustrated in the problem.

 

[Math10]

Is that the right way to use exact arithmetic to solve the system? No augmented matrix and Gauss-Jordan method involved?

 

[Ray Vickson]

Is that the right way to use exact arithmetic to solve the system? No augmented matrix and Gauss-Jordan method involved?

No, you can use whatever method you want (Gaussian elimination, for example); it is just that instead of using decimal numbers in the procedure you use rationals throughout. For example, consider the system
[tex] 6 x + 8 y = 19\\
15 x - 9y = 12[/tex]
The augmented form is
[tex] \begin{array}{rr|r}
6 & 8 & 19\\
15 & -9 & 12
\end{array}[/tex]
After one stop of Gaussian elimination we have
[tex] \begin{array}{rr|r}
6 & 8 & 19\\
0 & -29 & -71/2
\end{array}[/tex]
Thus,
[tex]0 x - 29 y = -71/2 \: \Longrightarrow y = (-71/2)/(-29) = 71/58.[/tex] Also,
[tex] x = (1/6)[19 - 8y] \; \Longrightarrow x = (1/6)[19 - 8(71/58)] = 89/58.[/tex]

 

[Math10]

What do you mean to use rationals? Do you meant that I convert 10^-3 into a fraction?

 

[Math10]

So I set up the augmented matrix and I want to solve this system using Gauss-Jordan method.

1/1000 -1
1 1

Is this how I solve this problem with 1 and 0 after the straight line in the right?

 

[verty]

So I set up the augmented matrix and I want to solve this system using Gauss-Jordan method.

1/1000 -1
1 1

Is this how I solve this problem with 1 and 0 after the straight line in the right?

Do you know the matrix method? If so, use it to solve it with those numbers.