xcopyv4n

2023-02-28

Determine the value of k so that the following linear equations have no solution:$(3k+1)x+3y-2=0\phantom{\rule{0ex}{0ex}}({k}^{2}+1)x+(k-2)y-5=0$

Nhluvukoj6m

Beginner2023-03-01Added 6 answers

The given equational system is

(3k + 1) x + 3y − 2 = 0

$({k}^{2}+1)x+(k-2)y-5=0$

Here, a1 = 3k + 1, b1 = 3, c1 = −2

a2 = k2 + 1, b2 = k − 2, c2 = −5

There is no answer to the given system of equations.

$\therefore \frac{a}{1}=\frac{b}{1}\ne \frac{c}{1}\Rightarrow \frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\ne \frac{-2}{-5}\Rightarrow \frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\mathrm{and}\frac{3}{k-2}\ne \frac{2}{5}$

Now,

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

$\Rightarrow k=-1$

When k = −1,

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

Thus, for k = −1, $\frac{3}{k-2}\ne \frac{2}{5}$

Therefore, the given system of equations has no solution when k = −1.

(3k + 1) x + 3y − 2 = 0

$({k}^{2}+1)x+(k-2)y-5=0$

Here, a1 = 3k + 1, b1 = 3, c1 = −2

a2 = k2 + 1, b2 = k − 2, c2 = −5

There is no answer to the given system of equations.

$\therefore \frac{a}{1}=\frac{b}{1}\ne \frac{c}{1}\Rightarrow \frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\ne \frac{-2}{-5}\Rightarrow \frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\mathrm{and}\frac{3}{k-2}\ne \frac{2}{5}$

Now,

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

$\Rightarrow k=-1$

When k = −1,

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

Thus, for k = −1, $\frac{3}{k-2}\ne \frac{2}{5}$

Therefore, the given system of equations has no solution when k = −1.