Chapter 4 Linear Equations In Two Variables

We need to write each given linear equation in the standard form $ax + by + c = 0$ and identify the coefficients $a$, $b$, and the constant term $c$.

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(i) $2x + 3y = 4.37$

To get the form $ax + by + c = 0$, we need to move the constant term to the left side of the equation.

$2x + 3y - 4.37 = 0$

Comparing this with $ax + by + c = 0$, we have:

$a = 2$

$b = 3$

$c = -4.37$

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(ii) $x - 4 = \sqrt{3} y$

Move all terms to the left side of the equation.

$x - \sqrt{3}y - 4 = 0$

Comparing this with $ax + by + c = 0$, we have:

$a = 1$

$b = -\sqrt{3}$

$c = -4$

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(iii) $4 = 5x - 3y$

Move all terms to one side. It is common practice to move terms to the left side or arrange it such that the coefficient of $x$ is positive.

Option 1 (Move RHS to LHS):

$4 - 5x + 3y = 0$

Rearranging in the form $ax + by + c = 0$:

$-5x + 3y + 4 = 0$

Comparing with $ax + by + c = 0$:

$a = -5$

$b = 3$

$c = 4$

Option 2 (Move LHS to RHS and flip the equation):

$0 = 5x - 3y - 4$

Flipping the equation:

$5x - 3y - 4 = 0$

Comparing with $ax + by + c = 0$:

$a = 5$

$b = -3$

$c = -4$

Both sets of values for $a$, $b$, and $c$ are valid, as multiplying the equation by -1 results in an equivalent equation.

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(iv) $2x = y$

Move the term $y$ to the left side of the equation.

$2x - y = 0$

To fit the form $ax + by + c = 0$, we can write it as:

$2x - 1y + 0 = 0$

Comparing this with $ax + by + c = 0$, we have:

$a = 2$

$b = -1$

$c = 0$

To write a linear equation in one variable as an equation in two variables ($x$ and $y$), we can include the missing variable with a coefficient of zero. The standard form of a linear equation in two variables is $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero.

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(i) $x = -5$

This equation involves only the variable $x$. We can include the variable $y$ with a coefficient of 0.

Move the constant term to the left side:

$x + 5 = 0$

Include the $y$ term with a coefficient of 0:

$1 \cdot x + 0 \cdot y + 5 = 0$

So, the equation in two variables is $x + 0y + 5 = 0$.

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(ii) $y = 2$

This equation involves only the variable $y$. We can include the variable $x$ with a coefficient of 0.

Move the constant term to the left side:

$y - 2 = 0$

Include the $x$ term with a coefficient of 0:

$0 \cdot x + 1 \cdot y - 2 = 0$

So, the equation in two variables is $0x + y - 2 = 0$.

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(iii) $2x = 3$

This equation involves only the variable $x$. Move the constant term to the left side:

$2x - 3 = 0$

Include the $y$ term with a coefficient of 0:

$2x + 0 \cdot y - 3 = 0$

So, the equation in two variables is $2x + 0y - 3 = 0$.

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(iv) $5y = 2$

This equation involves only the variable $y$. Move the constant term to the left side:

$5y - 2 = 0$

Include the $x$ term with a coefficient of 0:

$0 \cdot x + 5y - 2 = 0$

So, the equation in two variables is $0x + 5y - 2 = 0$.

Given:

Cost of a notebook = $\textsf{₹} x$

Cost of a pen = $\textsf{₹} y$

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

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To Represent:

Write a linear equation in two variables representing the given statement.

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Solution:

According to the statement, the cost of a notebook is equal to two times the cost of a pen.

Mathematically, this can be written as:

Cost of notebook = 2 $\times$ Cost of pen

Substitute the given variables:

$x = 2 \times y$

$x = 2y$

To write this equation in the standard form of a linear equation in two variables, which is $ax + by + c = 0$, we move all terms to one side of the equation.

Subtract $2y$ from both sides:

$x - 2y = 2y - 2y$

$x - 2y = 0$

This can be written as $1 \cdot x + (-2) \cdot y + 0 = 0$.

The linear equation in two variables representing the given statement is:

$x - 2y = 0$

We need to express each linear equation in the standard form $ax + by + c = 0$ and determine the values of $a$, $b$, and $c$ for each case.

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(i) $2x + 3y = 9.3\overline{5}$

To express this in the form $ax + by + c = 0$, we transpose the constant term to the left side:

$2x + 3y - 9.3\overline{5} = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 2$

$b = 3$

$c = -9.3\overline{5}$

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

This equation is already in the form $ax + by + c = 0$. We can rewrite the $\frac{y}{5}$ term as $\frac{1}{5}y$ or $1 \cdot x + (-\frac{1}{5})y - 10 = 0$.

$1x - \frac{1}{5}y - 10 = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 1$

$b = -\frac{1}{5}$

$c = -10$

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

To express this in the form $ax + by + c = 0$, we transpose the constant term to the left side:

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

Comparing this with $ax + by + c = 0$, we get:

$a = -2$

$b = 3$

$c = -6$

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(iv) $x = 3y$

To express this in the form $ax + by + c = 0$, we transpose the term $3y$ to the left side:

$x - 3y = 0$

We can write this as:

$1x - 3y + 0 = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 1$

$b = -3$

$c = 0$

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(v) $2x = -5y$

To express this in the form $ax + by + c = 0$, we transpose the term $-5y$ to the left side:

$2x + 5y = 0$

We can write this as:

$2x + 5y + 0 = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 2$

$b = 5$

$c = 0$

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(vi) $3x + 2 = 0$

This equation involves only $x$ and a constant. To express it in the form $ax + by + c = 0$, we include the $y$ term with a coefficient of 0:

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

Comparing this with $ax + by + c = 0$, we get:

$a = 3$

$b = 0$

$c = 2$

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(vii) $y - 2 = 0$

This equation involves only $y$ and a constant. To express it in the form $ax + by + c = 0$, we include the $x$ term with a coefficient of 0:

$0 \cdot x + 1 \cdot y - 2 = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 0$

$b = 1$

$c = -2$

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(viii) $5 = 2x$

To express this in the form $ax + by + c = 0$, we can rearrange the equation as $2x = 5$ and then transpose the constant term to the left side:

$2x - 5 = 0$

We can write this as:

$2x + 0 \cdot y - 5 = 0$

Comparing this with $ax + by + c = 0$, we get:

$a = 2$

$b = 0$

$c = -5$

Given:

The linear equation in two variables is $x + 2y = 6$.

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To Find:

Four different solutions $(x, y)$ that satisfy the equation.

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Solution:

A solution to a linear equation in two variables is a pair of values $(x, y)$ that makes the equation true. We can find solutions by choosing a value for one variable and then solving for the other variable.

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Finding Solution 1:

Let's choose $x = 0$. Substitute this value into the equation:

$0 + 2y = 6$

$2y = 6$

Divide both sides by 2:

$\frac{2y}{2} = \frac{6}{2}$

$y = 3$

So, one solution is $(x, y) = (0, 3)$.

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Finding Solution 2:

Let's choose $y = 0$. Substitute this value into the equation:

$x + 2(0) = 6$

$x + 0 = 6$

$x = 6$

So, another solution is $(x, y) = (6, 0)$.

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Finding Solution 3:

Let's choose $x = 2$. Substitute this value into the equation:

$2 + 2y = 6$

Subtract 2 from both sides:

$2y = 6 - 2$

$2y = 4$

Divide both sides by 2:

$\frac{2y}{2} = \frac{4}{2}$

$y = 2$

So, a third solution is $(x, y) = (2, 2)$.

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Finding Solution 4:

Let's choose $y = 1$. Substitute this value into the equation:

$x + 2(1) = 6$

$x + 2 = 6$

Subtract 2 from both sides:

$x = 6 - 2$

$x = 4$

So, a fourth solution is $(x, y) = (4, 1)$.

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Thus, four different solutions for the equation $x + 2y = 6$ are $(0, 3)$, $(6, 0)$, $(2, 2)$, and $(4, 1)$.

To find two different solutions for each linear equation, we can choose a value for one variable and solve for the other, or rearrange the equation to express one variable in terms of the other and then pick values.

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(i) $4x + 3y = 12$

Let's find the solutions by setting one variable to a specific value.

Solution 1:

Let $x = 0$. Substitute this into the equation:

$4(0) + 3y = 12$

$0 + 3y = 12$

$3y = 12$

Divide by 3:

$y = \frac{12}{3} = 4$

So, $(0, 4)$ is a solution.

Solution 2:

Let $y = 0$. Substitute this into the equation:

$4x + 3(0) = 12$

$4x + 0 = 12$

$4x = 12$

Divide by 4:

$x = \frac{12}{4} = 3$

So, $(3, 0)$ is another solution.

Two solutions for $4x + 3y = 12$ are $(0, 4)$ and $(3, 0)$.

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(ii) $2x + 5y = 0$

Solution 1:

Let $x = 0$. Substitute this into the equation:

$2(0) + 5y = 0$

$0 + 5y = 0$

$5y = 0$

Divide by 5:

$y = \frac{0}{5} = 0$

So, $(0, 0)$ is a solution.

Solution 2:

Let's choose another value, since setting $y=0$ will also give $x=0$. Let $x = 5$. Substitute this into the equation:

$2(5) + 5y = 0$

$10 + 5y = 0$

Subtract 10 from both sides:

$5y = -10$

Divide by 5:

$y = \frac{-10}{5} = -2$

So, $(5, -2)$ is another solution.

Two solutions for $2x + 5y = 0$ are $(0, 0)$ and $(5, -2)$.

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(iii) $3y + 4 = 0$

This equation can be written as $0x + 3y + 4 = 0$. This means the value of $x$ can be any real number, but the value of $y$ is fixed by the equation.

Solve for $y$:

$3y = -4$

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

So, any point with a y-coordinate of $-\frac{4}{3}$ is a solution. We can choose any two different values for $x$.

Solution 1:

Let $x = 0$. Since the value of $y$ is always $-\frac{4}{3}$, the solution is $(0, -\frac{4}{3})$.

Solution 2:

Let $x = 3$. Since the value of $y$ is always $-\frac{4}{3}$, the solution is $(3, -\frac{4}{3})$.

Two solutions for $3y + 4 = 0$ are $(0, -\frac{4}{3})$ and $(3, -\frac{4}{3})$.

The given equation is $y = 3x + 5$. This is a linear equation in two variables ($x$ and $y$) because it can be written in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers and $a$ and $b$ are not both zero.

Rearranging the equation $y = 3x + 5$ to the standard form:

$-3x + 1y - 5 = 0$

Here, $a = -3$, $b = 1$, and $c = -5$. Since $a = -3 \neq 0$ and $b = 1 \neq 0$, this is indeed a linear equation in two variables.

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A solution to a linear equation in two variables $(x, y)$ is a pair of values $(x, y)$ that satisfies the equation.

Let's see if we can find multiple solutions:

If $x = 0$, $y = 3(0) + 5 = 0 + 5 = 5$. Solution: $(0, 5)$.

If $x = 1$, $y = 3(1) + 5 = 3 + 5 = 8$. Solution: $(1, 8)$.

If $x = -1$, $y = 3(-1) + 5 = -3 + 5 = 2$. Solution: $(-1, 2)$.

If $x = 10$, $y = 3(10) + 5 = 30 + 5 = 35$. Solution: $(10, 35)$.

If we choose any real value for $x$, we can find a corresponding unique real value for $y$ by substituting the value of $x$ into the equation $y = 3x + 5$. Since there are infinitely many real numbers that can be chosen for $x$, there will be infinitely many corresponding values for $y$. Each such pair $(x, y)$ is a solution to the equation.

Graphically, a linear equation in two variables represents a straight line in the Cartesian plane. Every point on this line is a solution to the equation. A straight line consists of infinitely many points.

Therefore, the equation $y = 3x + 5$ has infinitely many solutions.

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The correct option is (iii) infinitely many solutions.

Reason: A linear equation in two variables of the form $ax + by + c = 0$, where $a$ and $b$ are not both zero, represents a straight line in the Cartesian plane. A straight line is composed of infinitely many points, and each point on the line corresponds to a solution $(x, y)$ that satisfies the equation. Thus, such an equation has infinitely many solutions.

To find four solutions for each linear equation, we can choose arbitrary values for one variable (either $x$ or $y$) and then solve for the corresponding value of the other variable.

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(i) $2x + y = 7$

We can rewrite the equation to express $y$ in terms of $x$: $y = 7 - 2x$. This makes it easier to find $y$ for chosen values of $x$.

Solution 1: Let $x = 0$.

$y = 7 - 2(0) = 7 - 0 = 7$

Solution: $(0, 7)$

Solution 2: Let $x = 1$.

$y = 7 - 2(1) = 7 - 2 = 5$

Solution: $(1, 5)$

Solution 3: Let $x = 2$.

$y = 7 - 2(2) = 7 - 4 = 3$

Solution: $(2, 3)$

Solution 4: Let $x = -1$.

$y = 7 - 2(-1) = 7 + 2 = 9$

Solution: $(-1, 9)$

Four solutions for $2x + y = 7$ are $(0, 7)$, $(1, 5)$, $(2, 3)$, and $(-1, 9)$.

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(ii) $\pi x + y = 9$

We can rewrite the equation to express $y$ in terms of $x$: $y = 9 - \pi x$.

Solution 1: Let $x = 0$.

$y = 9 - \pi(0) = 9 - 0 = 9$

Solution: $(0, 9)$

Solution 2: Let $x = 1$.

$y = 9 - \pi(1) = 9 - \pi$

Solution: $(1, 9 - \pi)$

Solution 3: Let $x = 2$.

$y = 9 - \pi(2) = 9 - 2\pi$

Solution: $(2, 9 - 2\pi)$

Solution 4: Let $x = \frac{9}{\pi}$.

$y = 9 - \pi(\frac{9}{\pi}) = 9 - 9 = 0$

Solution: $(\frac{9}{\pi}, 0)$

Four solutions for $\pi x + y = 9$ are $(0, 9)$, $(1, 9 - \pi)$, $(2, 9 - 2\pi)$, and $(\frac{9}{\pi}, 0)$.

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(iii) $x = 4y$

This equation already expresses $x$ in terms of $y$. It's easier to choose values for $y$ and find the corresponding values of $x$.

Solution 1: Let $y = 0$.

$x = 4(0) = 0$

Solution: $(0, 0)$

Solution 2: Let $y = 1$.

$x = 4(1) = 4$

Solution: $(4, 1)$

Solution 3: Let $y = -1$.

$x = 4(-1) = -4$

Solution: $(-4, -1)$

Solution 4: Let $y = 2$.

$x = 4(2) = 8$

Solution: $(8, 2)$

Four solutions for $x = 4y$ are $(0, 0)$, $(4, 1)$, $(-4, -1)$, and $(8, 2)$.

To check if a given ordered pair $(x, y)$ is a solution of the equation $x - 2y = 4$, we substitute the values of $x$ and $y$ into the equation and check if the left-hand side (LHS) is equal to the right-hand side (RHS), which is 4.

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(i) (0, 2)

Substitute $x = 0$ and $y = 2$ into the equation $x - 2y = 4$:

LHS $= 0 - 2(2)$

LHS $= 0 - 4$

LHS $= -4$

RHS $= 4$

Since LHS $\neq$ RHS ($-4 \neq 4$), $(0, 2)$ is not a solution.

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(ii) (2, 0)

Substitute $x = 2$ and $y = 0$ into the equation $x - 2y = 4$:

LHS $= 2 - 2(0)$

LHS $= 2 - 0$

LHS $= 2$

RHS $= 4$

Since LHS $\neq$ RHS ($2 \neq 4$), $(2, 0)$ is not a solution.

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(iii) (4, 0)

Substitute $x = 4$ and $y = 0$ into the equation $x - 2y = 4$:

LHS $= 4 - 2(0)$

LHS $= 4 - 0$

LHS $= 4$

RHS $= 4$

Since LHS $=$ RHS ($4 = 4$), $(4, 0)$ is a solution.

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(iv) $(\sqrt{2} , 4\sqrt{2})$

Substitute $x = \sqrt{2}$ and $y = 4\sqrt{2}$ into the equation $x - 2y = 4$:

LHS $= \sqrt{2} - 2(4\sqrt{2})$

LHS $= \sqrt{2} - 8\sqrt{2}$

LHS $= (1 - 8)\sqrt{2}$

LHS $= -7\sqrt{2}$

RHS $= 4$

Since LHS $\neq$ RHS ($-7\sqrt{2} \neq 4$), $(\sqrt{2}, 4\sqrt{2})$ is not a solution.

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(v) (1, 1)

Substitute $x = 1$ and $y = 1$ into the equation $x - 2y = 4$:

LHS $= 1 - 2(1)$

LHS $= 1 - 2$

LHS $= -1$

RHS $= 4$

Since LHS $\neq$ RHS ($-1 \neq 4$), $(1, 1)$ is not a solution.

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In summary:

• (0, 2) is not a solution.
• (2, 0) is not a solution.
• (4, 0) is a solution.
• $(\sqrt{2} , 4\sqrt{2})$ is not a solution.
• (1, 1) is not a solution.

Given:

The equation is $2x + 3y = k$.

We are given that $(x, y) = (2, 1)$ is a solution to this equation.

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To Find:

The value of $k$.

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Solution:

Since $(x, y) = (2, 1)$ is a solution to the equation $2x + 3y = k$, substituting these values for $x$ and $y$ into the equation must satisfy it.

Substitute $x = 2$ and $y = 1$ into the equation:

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

Perform the multiplication:

$4 + 3 = k$

Perform the addition:

$7 = k$

So, the value of $k$ is 7.

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The value of $k$ is 7.