We are all aware of the usual graphical method of solving a multi-variable system of inequalities. But how could one solve such a system algebraically? Also, does a system of inequalities follow the usual arithmetic of simultaneous equations? For instance, does the system \[ x+y>3 \\ x+2y>5 \]

necessarily imply \(y>2\). ( It does not seem to, since \(x=100\) and \(y=1\) is a perfectly valid solution)

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## Comments

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TopNewestNo, algebra of inequalities is not that simple... You cannot just simply subtract or divide two inequalities like we do to two equations. As its evident that you have subtracted inequalitities which isn't allowed and hence gave wrong results..... As an explicit example, if it had been:

\[x+y>3~~\text{and}~~ x+2y<5\] Then multiply first equation by \((-1)\) and add to second equation to get:

\[-(\not x+y)+\not x+2y<2\iff y<2\]

Note:- You can only add two inequalities together. You will not always be able to multiply, subtract or divide them.

PS:- I don't know any algebraic method to solve system these type of linear equations in two variables as I always relied on graph for getting solutions and they did the job fairly easily!

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Note: You cannot multiply two inequalities. For example, \( -2 < 1 \) and \( -2 < 1 \). Can we multiply them to conclude that \( 4 < 1 \)?

What is true is that you can multiply

anddivide by terms with the same sign. IE You can multiply/divide by \( - (x+y)^2 \), but not by \( -x+y \).Log in to reply

True that.....

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So, your note should be

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@Calvin Lin

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@Rishabh Cool

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@Nihar Mahajan

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If I've understood what you mean I think you could divide this system in 3 disequalities (a>b>c --> a>b, b>c, a>c) but I don't know if it works. Anyway with the values x=100 and y=1 you obtain 101 > 302 > 5, that is not true.

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Sorry for the typo. I have edited the note

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