Question

The number of value of k, for which the system of equations: (k+1)x+8y=4k; kx+(k+3)y=3k-1; has no solution, is:
(a) 1
(b) 2
(c) 3
(d) infinite

Answer 1

Hint:First, solve the two equations to get the value of x and y in terms of k. Then justify that for what values of k the variables x and y are not real.

Complete step-by-step answer:
Let us first write the equations that are given in the question.
The first equation given in the question is:
(k+1)x+8y=4k………………(i)
The second equation which is given in the question is:
kx+(k+3)y=3k-1…………..(ii)
Now we will multiply equation (i) by (k) and subtract it from equation (ii) multiplied with (k+1).

& \text{ }k\left( k+1 \right)x+\left( k+1 \right)\left( k+3 \right)y=\left( k+1 \right)\left( 3k-1 \right)\text{ } \\\ & \text{ }k\left( k+1 \right)x+8ky=4{{k}^{2}} \\\ & \underline{\text{- - - }} \\\ & \left( \left( k+1 \right)\left( k+3 \right)-8k \right)y=\left( k+1 \right)\left( 3k-1 \right)-4{{k}^{2}} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \left( {{k}^{2}}+4k+3-8k \right)y=3{{k}^{2}}+2k-1-4{{k}^{2}} \\\ & \Rightarrow y=\dfrac{-{{k}^{2}}+2k-1}{{{k}^{2}}-4k+3}=\dfrac{-{{\left( k-1 \right)}^{2}}}{\left( k-3 \right)\left( k-1 \right)}=\dfrac{-\left( k-1 \right)}{k-3} \\\ \end{aligned}$$ Now we will substitute the value of y in equation (i). On doing so, we get $$\left( k+1 \right)x-\dfrac{8\left( k-1 \right)}{k-3}=4k$$ $$\Rightarrow x=\dfrac{\dfrac{8\left( k-1 \right)}{k-3}+4k}{\left( k+1 \right)}$$ $$\Rightarrow x=\dfrac{8k-8+4{{k}^{2}}-12k}{\left( k+1 \right)\left( k-3 \right)}=\dfrac{4{{k}^{2}}-4k-8}{\left( k+1 \right)\left( k-3 \right)}$$ As we can see that k must not be equal to -1 and 3 to ensure that the denominator of x or y is not equal to zero. So, there are two values of k for which the system of equations is not defined. Hence, the answer to the above question is option (b). Note: The question above could have been solved using Cramer’s rule as well, but it is preferred to use Cramer’s rule when the number of variables is more than 2, and you are not able to find all the variables in terms of a single element. Also, be careful about the solving part, as it might be complicated and lead to mistakes.