Finding the Value of k for No Solution in Linear Equations

We are given a system of two linear equations in two variables x and y:

Equation 1: $(3k + 1)x + 3y - 2 = 0$

Equation 2: $(k^2 + 1)x + (k - 2)y - 5 = 0$

These equations are in the standard form $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.

Comparing the given equations with the standard form, we can identify the coefficients:

• From Equation 1: $a_1 = 3k + 1$, $b_1 = 3$, $c_1 = -2$
• From Equation 2: $a_2 = k^2 + 1$, $b_2 = k - 2$, $c_2 = -5$

Condition for No Solution

A system of linear equations has no solution if the lines represented by the equations are parallel and distinct. This condition is mathematically expressed as:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

Applying the Condition to Find k

Let's apply the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ to the given coefficients:

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

To solve for k, we can cross-multiply:

$(3k + 1)(k - 2) = 3(k^2 + 1)$

Expand both sides:

$3k(k - 2) + 1(k - 2) = 3k^2 + 3$

$3k^2 - 6k + k - 2 = 3k^2 + 3$

$3k^2 - 5k - 2 = 3k^2 + 3$

Subtract $3k^2$ from both sides:

$-5k - 2 = 3$

Add 2 to both sides:

$-5k = 3 + 2$

$-5k = 5$

Divide by -5:

$k = \frac{5}{-5}$

$k = -1$

Checking the Inequality Condition

Now we need to check if for $k = -1$, the condition $\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ is satisfied. Let's calculate the ratios for $k = -1$:

• $\frac{a_1}{a_2} = \frac{3(-1) + 1}{(-1)^2 + 1} = \frac{-3 + 1}{1 + 1} = \frac{-2}{2} = -1$
• $\frac{b_1}{b_2} = \frac{3}{-1 - 2} = \frac{3}{-3} = -1$
• $\frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5}$

For $k = -1$, we have $\frac{a_1}{a_2} = -1$, $\frac{b_1}{b_2} = -1$, and $\frac{c_1}{c_2} = \frac{2}{5}$.

We observe that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = -1$ and $\frac{c_1}{c_2} = \frac{2}{5}$.

Since $-1 \neq \frac{2}{5}$, the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ is true for $k = -1$.

Conclusion

The value of k for which the given system of linear equations has no solution is $k = -1$. This is because for $k=-1$, the ratios of the coefficients of x and y are equal ($\frac{a_1}{a_2} = \frac{b_1}{b_2}$), but this common ratio is not equal to the ratio of the constant terms ($\neq \frac{c_1}{c_2}$).