# System of linear equations with fractions

1. Feb 4, 2014

### Chase

1. The problem statement, all variables and given/known data
Solve the following system of linear equations.

$\begin{array}{cc} \frac{1}{2}x+y-\frac{3}{4}z=1 \\ \frac{2}{3}x-\frac{1}{3}y+z=2 \\ x-\frac{1}{5}y+2z=1 \end{array}$

3. The attempt at a solution
Can I just do elimination by addition? So if I multiple the first equation by $\frac{1}{3}$ I get $\frac{1}{3}\left(\frac{1}{2}x\right)+\frac{1}{3}y+\frac{1}{3}\left(-\frac{3}{4}\right)=\frac{1}{3}$ then I can add this to the second equation to eliminate the y's?

2. Feb 4, 2014

### SteamKing

Staff Emeritus
Yep, elimination works with fractions just like it does for whole number coefficients. It also works with decimal coefficients as well, just not as neatly.

3. Feb 4, 2014

### Chase

In my book it suggests to try clearing the denominators first to make it easier. To do this do you just multiply each term by the lowest commom number? So in my case I would multiply the first term by 2 to clear the 2, the second term by 4 giving 4y, leaving the third term as it is and then finally the 1 giving 4?

4. Feb 4, 2014

### Dick

You don't multiply every term by a different number. You multiply every equation by a different number. Multiply the first equation by 4, then second by 3 and the third by 5. Do you see why?

5. Feb 4, 2014

### Chase

Yes, to clear the fractions. When I multiply each equation as you said I'll end up with

\begin{array}{cc} 2x+4y-3z=4

\\ 2x-y+3z=6

\\ 5x-y+10z=5 \end{array}

?

Last edited: Feb 4, 2014
6. Feb 4, 2014

### Staff: Mentor

Looks fine.

BTW, using LaTeX here is overkill. You can write it in plain old text.

2x+4y-3z=4
2x-y+3z=6
5x-y+10z=5

7. Feb 5, 2014

### Ray Vickson

This is correct. However, clearing fractions like that may be a big waste of time, since as soon as you start to solve the equations you will be back to fractions again.