Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

3 Answers 3

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 $\LaTeX$

share|improve this answer
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.