# Let c be a constant. For what values of c can the simultaneous equations x-y=2; cx+y=3 have a solution (x, y) inside quadrant l?

Nov 24, 2016

In the first quadrant, both $x$ values and $y$ values are positive.

$\left\{\begin{matrix}- y = 2 - x \\ y = 3 - c x\end{matrix}\right.$

$- \left(3 - c x\right) = 2 - x$

$- 3 + c x = 2 - x$

$c x + x = 5$

$x \left(c + 1\right) = 5$

$x = \frac{5}{c + 1}$

We need $x > 0$ for there to be a solution in quadrant $1$.

$\frac{5}{c + 1} > 0$

There will be a vertical asymptote at $c = - 1$. Pick test points to the left and to the right of this asymptote.

Let $c = - 2$ and $c = 2$.

$\frac{5}{3 \left(- 2\right) + 1} = \frac{5}{- 5} = - 1$

$\therefore - 1 {>}^{\emptyset} 0$

So, the solution is $c > - 1$.

Hence, all values of $c$ that are larger than $- 1$ will ensure that the intersection points are in the first quadrant.

Hopefully this helps!

Nov 25, 2016

$- \frac{3}{2} < c < 1$

#### Explanation:

The equation $x - y = 2 \Leftrightarrow y = x - 2$ and hence this represents a line whose slope is $1$ and intercept on $y$-axis is $- 2$. Also intercept on $x$-axis can be obtained by putting $y = 0$ and is $2$. Equation of line appears as follows:
graph{x-2 [-10, 10, -5, 5]}

The other equation is $c x + y = 3$ or $y = - c x + 3$, which represents a line with $y$ intercept and slope $- c$. For this line to intersect above line in $Q 1$,

(i) it should have a minimum slope that of line joining $\left(0 , 3\right)$ and intercept of above line on $x$-axis i.e. at $\left(2 , 0\right)$, which is $\frac{0 - 3}{2 - 0} = - \frac{3}{2}$

and (ii) it should be passing through $\left(3 , 0\right)$ but have slope not more than $1$, as it will then intersect the line $x - y = 2$ in $Q 3$.

Hence, values of $c$ for which simultaneous equations $x - y = 2$ and $c x + y = 3$ have a solution $\left(x , y\right)$ inside $Q 1$ are given by

$- \frac{3}{2} < c < 1$

graph{(x-y-2)(x-y+3)(3x+2y-6)=0 [-10, 10, -5, 5]}