Math worksheet on linear equations in two variables for Class 10th NCERT students.

Here are the step-by-step solutions for each question on your worksheet.

Q.1) Find the value of k for which the system of equations has no solution:

Equations:
1) $3x - 4y + 7 = 0$
2) $kx + 3y - 5 = 0$

Step 1: Identify the coefficients.
For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:
* From equation (1): $a_1 = 3$, $b_1 = -4$, $c_1 = 7$
* From equation (2): $a_2 = k$, $b_2 = 3$, $c_2 = -5$

Step 2: Apply the condition for "no solution".
For a system to have no solution (parallel lines), the ratio of the coefficients of $x$ and $y$ must be equal, but not equal to the ratio of the constants:
$$ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $$

Step 3: Solve for k.
We use the first part of the condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2}$

$$ \frac{3}{k} = \frac{-4}{3} $$

Cross-multiply to solve for $k$:
$$ 3 \times 3 = k \times (-4) $$
$$ 9 = -4k $$
$$ k = \frac{9}{-4} $$
$$ k = -\frac{9}{4} $$

*(Check: Does this satisfy the second part? $\frac{-4}{3} \neq \frac{7}{-5}$. Yes, they are not equal. So the answer is valid.)*

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Q.2) Give a linear equation $2x - 3y = 10$. Write another linear equation such that the lines represented by the pair are:

The given equation is $2x - 3y - 10 = 0$. Here, $a_1 = 2$ and $b_1 = -3$.

Part A: Intersecting lines
Condition: The ratios of the coefficients of $x$ and $y$ must NOT be equal ($\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$).
* We need an equation where the slope is different.
* Let's keep the $x$ coefficient as 2, but change the $y$ coefficient to something other than -3. Let's pick $+3y$.
* Equation: $2x + 3y = 5$
* Check: $\frac{2}{2} \neq \frac{-3}{3}$ ($1 \neq -1$). This works.

Part B: Coincident lines
Condition: All ratios must be equal ($\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$).
* The easiest way to get coincident lines is to multiply the entire original equation by any non-zero number.
* Let's multiply $2x - 3y = 10$ by 2.
* $2(2x) - 2(3y) = 2(10)$
* Equation: $4x - 6y = 20$

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Q.3) Write the number of solutions for the following pair of linear equations:

Equations:
1) $x + 2y - 8 = 0$
2) $2x + 4y + 16 = 0$

Step 1: Identify coefficients.
* $a_1 = 1, b_1 = 2, c_1 = -8$
* $a_2 = 2, b_2 = 4, c_2 = 16$

Step 2: Compare the ratios.
* Ratio of $x$: $\frac{a_1}{a_2} = \frac{1}{2}$
* Ratio of $y$: $\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$
* Ratio of constants: $\frac{c_1}{c_2} = \frac{-8}{16} = -\frac{1}{2}$

Step 3: Conclusion.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ ($\frac{1}{2} = \frac{1}{2} \neq -\frac{1}{2}$), the lines are parallel. Parallel lines never meet.

Therefore, there is No Solution.

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Q.4) Find the value of k for which the system of equations has unique solution:

Equations:
1) $x + ky = 0$ (Note: constant term $c_1 = 0$)
2) $2x - y = 0$ (Note: constant term $c_2 = 0$)

Step 1: Identify coefficients.
* $a_1 = 1, b_1 = k$
* $a_2 = 2, b_2 = -1$

Step 2: Apply the condition for "unique solution".
For a unique solution (intersecting lines), the ratio of $x$ coefficients must NOT equal the ratio of $y$ coefficients:
$$ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} $$

Step 3: Solve for k.
$$ \frac{1}{2} \neq \frac{k}{-1} $$

Multiply both sides by -1 to isolate $k$:
$$ -\frac{1}{2} \neq k $$

So, $k$ can be any real number except $-\frac{1}{2}$.

Final Answer:
Q.1) $k = -\frac{9}{4}$

Q.2)
* Intersecting: $2x + 3y = 5$ (Note: Many answers are possible, e.g., $x + y = 1$)
* Coincident: $4x - 6y = 20$

Q.3) No Solution

Q.4) $k \neq -\frac{1}{2}$