System of linear equations with fractions

[Chase]

Homework Statement

Solve the following system of linear equations.

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

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?

 

[SteamKing]

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.

 

[Chase]

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.

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?

 

[Dick]

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?

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?

 

[Chase]

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?

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}

?

 

[Mark44]

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}

?

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

 

[Ray Vickson]

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}

?

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.