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If a have a linear system of three equations in three unknowns but two of the three equations are multiples of each other, does it mean the linear system has infinitely many solutions? Why?

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    $\begingroup$ Not necessarily. You essentially just have two equations that matter then. Since they're in three variables each, you can think of them as representing planes, and two (different) planes can intersect either in a line (infinitely many solutions) or not at all. $\endgroup$ – themathandlanguagetutor Sep 8 '17 at 7:17
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Linear equations of 3 variables represent planes in a 3D space.

If you have two distinct planes in a 3D space they intersect along a line(if they aren't parallel). Now if you have a third plane then that would intersect with this line on a single point. Or if the line is parallel it might not.

Since you two equations in that are multiples of each other then that means that they represent the same plane in the 3D space. And two planes would always intersect at a line if they aren't parallel. Since a line has infinitely many points there are infinitely many solutions to this.

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  • $\begingroup$ You’re assuming that the remaining two independent equations are consistent. That’s not a given. $\endgroup$ – amd Sep 8 '17 at 8:17
  • $\begingroup$ Yes you are right it is assuming that the third equation is consistent with the first two. $\endgroup$ – Sonal_sqrt Sep 8 '17 at 8:23

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