# My problem: x 1 </msub> + 3 x 2 </msub> &#x2212;<!-- −

My problem:
${x}_{1}+3{x}_{2}-{x}_{3}+{x}_{4}+2{x}_{5}=2,$
$2{x}_{1}-{x}_{2}+{x}_{3}-{x}_{4}+{x}_{5}=-2,$
$4{x}_{1}+2{x}_{2}+2{x}_{3}-{x}_{4}-{x}_{5}=0.$
I thought that I could multiply 1 by equation 2, and then add that to equation 1, which would cancel both the ${x}_{3}$ and the ${x}_{4}$ factors for that line. Is that permissible?
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Raven Higgins
What you suggest is just fine. But doing what your book suggests is not a bad idea either. Since you have three equations and five unknowns, you will only be able to solve for three of the variables (at most) in terms of the other two anyway.

Semaj Christian
By the way it's more clear if you make a scale reduction starting from ${x}_{1}$, ie keep the first line, substitute the second line with "second line minus two times the first". And substitute the 3rd one with "3rd minus 4 times the first". You get a new system. Repeat and rinse with ${x}_{2}$