Question

The number of values of k for which the system of equations x+y=2 , kx+y=4 , x+ky=5 has atleast one solution?

Answer 1

Conditions

The number of values of $k$ for which the system of equations $x + y = 2$, $kx + y = 4$, $x + ky = 5$ has at least one solution?

Solution

We must construct a system of 3 equations, where there will be 3 variables: $x, y$ and $k$:

$$\left\{ \begin{array}{l} x + y = 2 \\ kx + y = 4 \\ x + ky = 5 \end{array} \right.$$ $$x = 2 - y$$

$3^{\text{rd}}$ equation minus $1^{\text{st}}$ give us:

$$(k - 1)y = 3$$ $$y = \frac{3}{k - 1}$$

Then, from the second equation:

$$k\left(2 - \frac{3}{k - 1}\right) + \frac{3}{k - 1} = 4$$

It's obvious, that $k = 1$ couldn't be a solution (because all 3 equations would have equal left sides, but the right sides wouldn't be equal). So, let's multiply on (k-1):

$$2k(k - 1) - 3k + 3 = 4(k - 1)$$ $$2k^2 - 2k - 3k + 3 - 4k + 4 = 0$$ $$2k^2 - 9k + 7 = 0$$ $$D = 81 - 4 \cdot 2 \cdot 7 = 81 - 56 = 25 > 0$$

On this point we can already answer our question – as the discriminant is positive – there are 2 different solutions for $k$, that's why the number of values of $k$, for which the system of equations has at least one solution is **TWO VALUES**

To make sure it, we can find 2 values of $k$:

$$k_{1,2} = \frac{9 \pm 5}{4} = \left[ \begin{array}{l} k = \frac{7}{2} \\ k = 1 \end{array} \right]$$

After that we will have 2 pairs of $x$ and $y$, for each $k$, which could be found by substitution $k$-value in these equations:

$$x = 2 - y$$ $$y = \frac {3}{k - 1}$$