Understanding Linear Equations with No Solution

A pair of linear equations in two variables represents two straight lines in a coordinate plane. These lines can be intersecting, parallel, or coincident. The number of solutions depends on the relationship between these lines.

For a pair of linear equations given by:

• $\small a_1x + b_1y + c_1 = 0$
• $\small a_2x + b_2y + c_2 = 0$

The conditions for the existence and nature of solutions are:

• If $\small \frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a single point, and there is a unique solution.
• If $\small \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident, and there are infinitely many solutions.
• If $\small \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and distinct, and there is no solution.

The question asks for the condition under which the given pair of equations has no solution. This means we need to apply the third condition: $\small \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Identifying Coefficients in the Given Pair of Equations

The given pair of linear equations is:

• Equation 1: $\small 2x - ky + 7 = 0$
• Equation 2: $\small 6x - 12y + 15 = 0$

Comparing these equations with the general form $\small ax + by + c = 0$, we can identify the coefficients:

• From Equation 1: $\small a_1 = 2$, $\small b_1 = -k$, $\small c_1 = 7$
• From Equation 2: $\small a_2 = 6$, $\small b_2 = -12$, $\small c_2 = 15$

Applying the Condition for No Solution

For the pair of equations to have no solution, the ratio of coefficients must satisfy $\small \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.

Let's substitute the identified coefficients into this condition:

$\small \frac{2}{6} = \frac{-k}{-12} \neq \frac{7}{15}$

Step 1: Solve the equality part to find the value of k

We consider the equality part of the condition:

$\small \frac{2}{6} = \frac{-k}{-12}$

Simplify the fraction on the left side:

$\small \frac{1}{3} = \frac{k}{12}$

Now, solve for k by cross-multiplication or by multiplying both sides by 12:

$\small k = \frac{1 \times 12}{3}$

$\small k = \frac{12}{3}$

$\small k = 4$

Step 2: Verify the inequality part with the obtained value of k

Now we need to check if this value of k satisfies the inequality part of the condition:

$\small \frac{-k}{-12} \neq \frac{7}{15}$

Substitute $\small k = 4$ into the expression $\small \frac{-k}{-12}$:

$\small \frac{-4}{-12} = \frac{4}{12} = \frac{1}{3}$

So, the inequality becomes:

$\small \frac{1}{3} \neq \frac{7}{15}$

To check this, we can cross-multiply:

$\small 1 \times 15 \neq 3 \times 7$

$\small 15 \neq 21$

This inequality is true. The ratio of the constant terms is indeed different from the ratio of the coefficients of x and y when $\small k=4$.

Conclusion: Value of k for No Solution

Since the value $\small k=4$ satisfies both the equality $\small \frac{a_1}{a_2} = \frac{b_1}{b_2}$ and the inequality $\small \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the pair of linear equations has no solution when $\small k=4$. This condition corresponds to the lines being parallel and distinct.

This confirms that for the given system of equations, the no solution case occurs at $\small k=4$. Finding the correct value of k is essential for solving such problems involving pairs of linear equations.