1. Find the value of k for no solution

Q1: Find the value of k for which the pair of linear equations 5x + 2y – 7 = 0 and 2x + ky + 1 = 0 don’t have a solution.[CBSE 2024]

(a) 5
(b) \[\frac{4}{5}\]
(c)\[\frac{5}{4}\]
(d)\[\frac{5}{2}\]
Solution:

For two equations to have no solution, the ratios of coefficients must satisfy:

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

For equations: 5x + 2y – 7 = 0 and 2x + ky + 1 = 0

\[ \frac{5}{2} = \frac{2}{k} \neq \frac{-7}{1} \]

Solving \[ \frac{5}{2} = \frac{2}{k} \]

\[ 5k = 4 \]

\[ k = \frac{4}{5} \]

Now check if \[ \frac{-7}{1} = -7 \neq \frac{5}{2} \]

Answer: (b) \(\frac{4}{5}\)

2. Solve the pair of linear equations

Q2: Solve the following pair of linear equations for x and y algebraically: x + 2y = 9 and y – 2x = 2.[CBSE 2024]

Solution:

Equation 1: x + 2y = 9

Equation 2: y – 2x = 2

Rewriting Equation 2: y = 2x + 2

Substitute into Equation 1:

x + 2(2x + 2) = 9

x + 4x + 4 = 9

5x + 4 = 9

5x = 5

x = 1

Substitute x = 1 into y = 2x + 2:

y = 2(1) + 2 = 4

Answer: x = 1, y = 4

3. Check if point lies on both lines

Q3: Check whether the point (-4, 3) lies on both the lines represented by the linear equations x + y + 1 = 0 and x – y = 1.[CBSE 2024]

Solution:

For equation 1: x + y + 1 = 0

Substitute x = -4, y = 3:

-4 + 3 + 1 = 0

0 = 0 ✓ (satisfied)

For equation 2: x – y = 1

Substitute x = -4, y = 3:

-4 – 3 = -7 ≠ 1 ✗ (not satisfied)

Answer: The point (-4, 3) lies on the first line (x + y + 1 = 0) but not on the second line (x – y = 1).

4. Consistency of linear equations from graph

Q4: In the given figure, graphs of two linear equations are shown. The pair of these linear equations is:

(a) consistent with unique solution
(b) consistent with infinitely many solutions
(c) inconsistent
(d) inconsistent but can be made consistent by extending these lines[CBSE 2024]

Solution:

From the graph, we can see that the two lines intersect at a single point. When two lines intersect at exactly one point, the system of equations is consistent and has a unique solution.

Answer: (a) consistent with unique solution