Question

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?

Answer 1

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

`{(-y = 2 - x), (y = 3 - cx):}`

`-(3 - cx) = 2 - x`

`-3 + cx = 2 - x`

`cx + x = 5`

`x(c + 1) = 5`

`x = 5/(c + 1)`

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

`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`.

`5/(3(-2) + 1) = 5/(-5)= -1`

`:. -1 >^O/ 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!

Answer 2

`-3/2 < c < 1`

Explanation:

The equation `x-y=2hArry=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 `cx+y=3` or `y=-cx+3`, which represents a line with `y` intercept and slope `-c`. For this line to intersect above line in `Q1`,

(i) it should have a minimum slope that of line joining `(0,3)` and intercept of above line on `x`-axis i.e. at `(2,0)`, which is `(0-3)/(2-0)=-3/2`

and (ii) it should be passing through `(3,0)` but have slope not more than `1`, as it will then intersect the line `x-y=2` in `Q3`.

Hence, values of `c` for which simultaneous equations `x-y=2` and `cx+y=3` have a solution `(x,y)` inside `Q1` are given by

`-3/2 < c < 1`

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