I have a system of linear inequalities such as:
$y\geq 2x+1$
$y>(x/2)-1$
I know how to solve this system of inequalities (in order to find possible values that will satisfy it) by graphing the two inequalities separately and finding values that will satisfy both simultaneously.
However, is it possible to solve the system algebraically in such a way that it will get another inequality that contains the possible solutions to the system, without having to resort to the graphing process?
$\endgroup$2 Answers
$\begingroup$You can compare the right-hand sides as$$2x+1\geq\frac{x}{2}-1$$ or $$\frac{x}{2}-1\geq 2x+1$$
$\endgroup$ 3 $\begingroup$If $$2x+1\geq\frac{x}{2}-1$$ or $$x\geq-\frac{4}{3}$$ we obtain:$$\left\{(x,y)|x\geq-\frac{4}{3},y\geq2x+1\right\}\setminus\left\{-\frac{4}{3},-\frac{5}{3}\right\}.$$If $$2x+1<\frac{x}{2}-1$$ or $$x<-\frac{4}{3}$$ we obtain:$$\left\{(x,y)|x<-\frac{4}{3},y>\frac{x}{2}-1\right\}.$$
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