Linear Equations in two Variables

Given: Cost of a notebook = ₹x, Cost of a pen = ₹y.

Statement: The cost of a notebook is twice the cost of a pen.

Forming the equation:
$$x = 2y$$

This can be written in standard form as:
$$x - 2y = 0$$

or equivalently $1 \cdot x + (-2) \cdot y + 0 = 0$.

Answer: The required linear equation in two variables is $x = 2y$ (or $x - 2y = 0$).

The standard form is $ax + by + c = 0$.

(i) $2x + 3y = 9.35$

Rewriting: $2x + 3y - 9.35 = 0$

Here, $a = 2,\; b = 3,\; c = -9.35$.

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(ii) $x - \dfrac{y}{5} - 10 = 0$

This is already in the required form: $1 \cdot x + \left(-\dfrac{1}{5}\right)y + (-10) = 0$

Here, $a = 1,\; b = -\dfrac{1}{5},\; c = -10$.

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(iii) $-2x + 3y = 6$

Rewriting: $-2x + 3y - 6 = 0$

Here, $a = -2,\; b = 3,\; c = -6$.

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(iv) $x = 3y$

Rewriting: $x - 3y = 0$, i.e., $1 \cdot x + (-3)y + 0 = 0$

Here, $a = 1,\; b = -3,\; c = 0$.

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(v) $2x = -5y$

Rewriting: $2x + 5y = 0$, i.e., $2x + 5y + 0 = 0$

Here, $a = 2,\; b = 5,\; c = 0$.

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(vi) $3x + 2 = 0$

Rewriting: $3x + 0 \cdot y + 2 = 0$

Here, $a = 3,\; b = 0,\; c = 2$.

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(vii) $y - 2 = 0$

Rewriting: $0 \cdot x + 1 \cdot y + (-2) = 0$

Here, $a = 0,\; b = 1,\; c = -2$.

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(viii) $5 = 2x$

Rewriting: $2x = 5 \Rightarrow 2x - 5 = 0$, i.e., $2x + 0 \cdot y + (-5) = 0$

Here, $a = 2,\; b = 0,\; c = -5$.

Correct option: (iii) infinitely many solutions.

Reason: The equation $y = 3x + 5$ is a linear equation in two variables $x$ and $y$. For every real value we assign to $x$, we get a corresponding real value of $y$. Since there are infinitely many real numbers to substitute for $x$, the equation has infinitely many solutions.

For example:
- $x = 0 \Rightarrow y = 5$ → solution $(0, 5)$
- $x = 1 \Rightarrow y = 8$ → solution $(1, 8)$
- $x = -1 \Rightarrow y = 2$ → solution $(-1, 2)$

Thus, a linear equation in two variables always has infinitely many solutions.

(i) $2x + y = 7$, so $y = 7 - 2x$.

| $x$ | $y = 7 - 2x$ | Solution |
|-----|--------------|----------|
| $0$ | $7$ | $(0,\ 7)$ |
| $1$ | $5$ | $(1,\ 5)$ |
| $2$ | $3$ | $(2,\ 3)$ |
| $3$ | $1$ | $(3,\ 1)$ |

Four solutions: $(0, 7),\ (1, 5),\ (2, 3),\ (3, 1)$.

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(ii) $\pi x + y = 9$, so $y = 9 - \pi x$.

| $x$ | $y = 9 - \pi x$ | Solution |
|-----|-----------------|----------|
| $0$ | $9$ | $(0,\ 9)$ |
| $1$ | $9 - \pi$ | $(1,\ 9-\pi)$ |
| $2$ | $9 - 2\pi$ | $(2,\ 9-2\pi)$ |
| $-1$ | $9 + \pi$ | $(-1,\ 9+\pi)$ |

Four solutions: $(0, 9),\ (1, 9-\pi),\ (2, 9-2\pi),\ (-1, 9+\pi)$.

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(iii) $x = 4y$, so $x = 4y$ (choose values of $y$).

| $y$ | $x = 4y$ | Solution |
|-----|----------|----------|
| $0$ | $0$ | $(0,\ 0)$ |
| $1$ | $4$ | $(4,\ 1)$ |
| $2$ | $8$ | $(8,\ 2)$ |
| $-1$ | $-4$ | $(-4,\ -1)$ |

Four solutions: $(0, 0),\ (4, 1),\ (8, 2),\ (-4, -1)$.

The equation is $x - 2y = 4$. We substitute each pair and check whether LHS $= 4$.

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(i) $(0, 2)$: $x = 0,\ y = 2$
$$\text{LHS} = 0 - 2(2) = 0 - 4 = -4 \neq 4$$
$\therefore (0, 2)$ is not a solution.

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(ii) $(2, 0)$: $x = 2,\ y = 0$
$$\text{LHS} = 2 - 2(0) = 2 - 0 = 2 \neq 4$$
$\therefore (2, 0)$ is not a solution.

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(iii) $(4, 0)$: $x = 4,\ y = 0$
$$\text{LHS} = 4 - 2(0) = 4 - 0 = 4 = \text{RHS}$$
$\therefore (4, 0)$ is a solution.

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(iv) $(\sqrt{2},\ 4\sqrt{2})$: $x = \sqrt{2},\ y = 4\sqrt{2}$
$$\text{LHS} = \sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = -7\sqrt{2} \neq 4$$
$\therefore (\sqrt{2},\ 4\sqrt{2})$ is not a solution.

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(v) $(1, 1)$: $x = 1,\ y = 1$
$$\text{LHS} = 1 - 2(1) = 1 - 2 = -1 \neq 4$$
$\therefore (1, 1)$ is not a solution.

Given: The equation is $2x + 3y = k$ and $(x, y) = (2, 1)$ is a solution.

Concept: If $(2, 1)$ is a solution, then substituting $x = 2$ and $y = 1$ must satisfy the equation.

Substituting:
$$2(2) + 3(1) = k$$
$$4 + 3 = k$$
$$k = 7$$

Answer: The value of $k$ is $\boxed{7}$.