xcopyv4n

2023-02-28

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

Nhluvukoj6m

The given equational system is
(3k + 1) x + 3y − 2 = 0
$\left({k}^{2}+1\right)x+\left(k-2\right)y-5=0$
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.
$\therefore \frac{a}{1}=\frac{b}{1}\ne \frac{c}{1}⇒\frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\ne \frac{-2}{-5}⇒\frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\mathrm{and}\frac{3}{k-2}\ne \frac{2}{5}$
Now,
$\frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}⇒\left(3k+1\right)\left(k-2\right)=3\left({k}^{2}+1\right)⇒3{k}^{2}-5k-2=3{k}^{2}+3⇒-5k=5$
$⇒k=-1$
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.

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