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Given the sytem of equation $y= -x+6$, $y= x/3+c$ with the solution lying in quadrant I, find all possible values of $c$.

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Write this as a system of linear equations, row reduce, we see $c$ can take only certain values if $x>0$ and $y>0$. Do it, it's not difficult but very troublesome to type using $\LaTeX$

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Since we're requiring that there be solutions in the first quadrant, then the equation $y=-x+6$ narrows down our possible $x$-values (and $y$-values) to the open interval $]0,6[$. (Why?) Solving the system of equations (I leave the steps to you) for $x$ gives us $$x=-\frac34 c+\frac92.$$ Now, solve the following for $c$ $$0<-\frac34 c+\frac 92<6,$$ then confirm that for each $c$ in the open interval thus determined, the corresponding system has a solution in the first quadrant.

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We sere obviously that $-x+6=x/3+c \implies 4x/3=6-c \implies x=\frac{3}{4}(6-c)$. Since $x>0$, we see $\frac{3}{4}(6-c)>0 \implies c<6$.

Solving instead in terms of $y$ gives $x=6-y \implies y=\frac{6-y}{3}+c \implies 4y/3=c+2$, giving $y=\frac{3}{4}(c+2)$. $y>0$ also so $c>-2$. Therefore $-2<c<6$.

At the endpoints of this interval, there are solutions on the co-ordinate axes. I do not know whether you want to include these or not.

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