Chapter 4 Linear Equations In Two Variables

The goal is to rearrange each equation into the standard form of a linear equation in two variables, which is $ax + by + c = 0$. In this form, 'a' is the coefficient of the x term, 'b' is the coefficient of the y term, and 'c' is the constant term, with all terms on one side of the equation and zero on the other.

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

To get the form $ax + by + c = 0$, we need to move the constant term (4.37) from the right side to the left side.

$2x + 3y - 4.37 = 0$

Now, we compare this equation term by term with the standard form $ax + by + c = 0$:

The term with x is $2x$, so $a = 2$.

The term with y is $+3y$, so $b = 3$.

The constant term is $-4.37$, so $c = -4.37$.

Therefore: $a = 2$, $b = 3$, $c = -4.37$.

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

To get the form $ax + by + c = 0$, we need to move the y term ($\sqrt{3} y$) from the right side to the left side.

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

Now, compare with $ax + by + c = 0$:

The term with x is $x$ (which is $1x$), so $a = 1$.

The term with y is $-\sqrt{3} y$, so $b = -\sqrt{3}$.

The constant term is $-4$, so $c = -4$.

Therefore: $a = 1$, $b = -\sqrt{3}$, $c = -4$.

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

To get the form $ax + by + c = 0$, we need to move all terms to one side.

$5x - 3y - 4 = 0$

Now, compare with $ax + by + c = 0$:

The term with x is $5x$, so $a = 5$.

The term with y is $-3y$, so $b = -3$.

The constant term is $-4$, so $c = -4$.

Therefore: $a = 5$, $b = -3$, $c = -4$.

(Alternatively, we could move the x and y terms to the left: $4 - 5x + 3y = 0$, which rearranges to $-5x + 3y + 4 = 0$. In this case, $a = -5$, $b = 3$, $c = 4$. Both forms are mathematically correct, but usually, we prefer the coefficient of x to be positive if possible.)

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(iv) $2x = y$

To get the form $ax + by + c = 0$, move the y term from the right side to the left side:

$2x - y = 0$

This equation doesn't explicitly show a constant term 'c'. We can think of it as having a constant term of zero:

$2x - 1y + 0 = 0$

Now, compare with $ax + by + c = 0$:

The term with x is $2x$, so $a = 2$.

The term with y is $-y$ (which is $-1y$), so $b = -1$.

The constant term is $0$, so $c = 0$.

Therefore: $a = 2$, $b = -1$, $c = 0$.

The standard form of a linear equation in two variables is $ax + by + c = 0$. To express an equation originally given with only one variable (like $x$ or $y$) in this two-variable form, we need to include the missing variable. We can do this by adding the missing variable multiplied by a coefficient of zero, since $0 \times (\text{variable}) = 0$, which doesn't change the original equation's meaning.

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(i) $x = -5$

First, rewrite the equation so one side is zero:

$x + 5 = 0$

This equation only involves 'x'. To include 'y' without changing the equation's solutions, we add a '0y' term:

$x + 0y + 5 = 0$

This can also be written explicitly showing the coefficient of x:

$1x + 0y + 5 = 0$

This is now in the form $ax + by + c = 0$ with $a=1$, $b=0$, $c=5$.

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(ii) $y = 2$

First, rewrite the equation so one side is zero:

$y - 2 = 0$

This equation only involves 'y'. To include 'x' without changing the equation, we add a '0x' term:

$0x + y - 2 = 0$

This can also be written explicitly showing the coefficient of y:

$0x + 1y - 2 = 0$

This is now in the form $ax + by + c = 0$ with $a=0$, $b=1$, $c=-2$.

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(iii) $2x = 3$

First, rewrite the equation so one side is zero:

$2x - 3 = 0$

This equation only involves 'x'. To include 'y', we add a '0y' term:

$2x + 0y - 3 = 0$

This is now in the form $ax + by + c = 0$ with $a=2$, $b=0$, $c=-3$.

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(iv) $5y = 2$

First, rewrite the equation so one side is zero:

$5y - 2 = 0$

This equation only involves 'y'. To include 'x', we add a '0x' term:

$0x + 5y - 2 = 0$

This is now in the form $ax + by + c = 0$ with $a=0$, $b=5$, $c=-2$.

We are asked to translate the given statement into a mathematical equation using two variables.

Step 1: Define the variables.

The problem specifies the variables to use:

• Let the cost of a notebook be ₹ x.
• Let the cost of a pen be ₹ y.

Step 2: Translate the statement into an equation.

The statement is: "The cost of a notebook is twice the cost of a pen."

Translating this using our variables:

(Cost of a notebook) = 2 $\times$ (Cost of a pen)

So, $x = 2y$

Step 3: Write the equation in the standard form $ax + by + c = 0$ (optional but good practice).

The standard form requires all terms to be on one side of the equation.

$x - 2y = 0$

This can also be written as $1x - 2y + 0 = 0$, explicitly showing the coefficients $a=1$, $b=-2$, and $c=0$.

The required linear equation in two variables representing the statement is $x - 2y = 0$ (or $x = 2y$).

We need to rearrange each equation into the standard form $ax + by + c = 0$ and then identify the coefficients $a$, $b$, and the constant term $c$.

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

Move the constant term to the left side by subtracting $9.3\overline{5}$ from both sides:

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

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

$a = 2$, $b = 3$, $c = -9.3\overline{5}$.

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

The equation is already in the form $ax + by + c = 0$. We need to identify the coefficients carefully.

The term with x is $x$, which is $1x$.

The term with y is $-\frac{y}{5}$, which can be written as $(-\frac{1}{5})y$.

The constant term is $-10$.

So, the equation can be seen as $1x + (-\frac{1}{5})y + (-10) = 0$.

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

$a = 1$, $b = -\frac{1}{5}$, $c = -10$.

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

Move the constant term to the left side by subtracting 6 from both sides:

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

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

$a = -2$, $b = 3$, $c = -6$.

Alternatively, we could multiply the entire equation by -1 to make the 'a' coefficient positive: $2x - 3y + 6 = 0$. In this case, $a = 2$, $b = -3$, $c = 6$. Both answers are correct representations.

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(iv) $x = 3y$

Move the y term to the left side by subtracting $3y$ from both sides:

$x - 3y = 0$.

To match the standard form $ax + by + c = 0$, we can explicitly write the coefficients and the constant term (which is 0 in this case):

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

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

$a = 1$, $b = -3$, $c = 0$.

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(v) $2x = -5y$

Move the y term to the left side by adding $5y$ to both sides:

$2x + 5y = 0$.

Explicitly writing the constant term:

$2x + 5y + 0 = 0$.

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

$a = 2$, $b = 5$, $c = 0$.

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(vi) $3x + 2 = 0$

This equation seems to have only one variable, x. To express it in the two-variable form $ax + by + c = 0$, we introduce the variable y with a coefficient of 0, since $0y = 0$.

$3x + 0y + 2 = 0$.

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

$a = 3$, $b = 0$, $c = 2$.

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(vii) $y - 2 = 0$

This equation seems to have only one variable, y. To express it in the two-variable form $ax + by + c = 0$, we introduce the variable x with a coefficient of 0, since $0x = 0$.

$0x + y - 2 = 0$.

We can write 'y' as '1y':

$0x + 1y - 2 = 0$.

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

$a = 0$, $b = 1$, $c = -2$.

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(viii) $5 = 2x$

First, rearrange to get all terms on one side. Subtract 5 from both sides:

$0 = 2x - 5$.

Or, swap sides: $2x - 5 = 0$.

This equation involves only x. Introduce the y variable with a coefficient of 0:

$2x + 0y - 5 = 0$.

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

$a = 2$, $b = 0$, $c = -5$.

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

A "solution" to this equation is a pair of values (one for x and one for y) that makes the equation true. Since it's a linear equation in two variables, it has infinitely many solutions. We need to find four distinct pairs (x, y).

A common strategy is to choose a value for one variable (say, y) and then solve the equation for the other variable (x). It can be helpful to first rearrange the equation to isolate one variable. Let's isolate x:

$x + 2y = 6$

Subtract 2y from both sides:

$x = 6 - 2y$

Now we can easily find x for any chosen value of y:

1. Choose $y = 0$:
   Substitute $y=0$ into $x = 6 - 2y$:
   $x = 6 - 2(0)$
   $x = 6 - 0$
   $x = 6$.
   So, when $y=0$, $x=6$. The solution is the ordered pair (6, 0).
   Check: $6 + 2(0) = 6 + 0 = 6$. (True)
2. Choose $y = 1$:
   Substitute $y=1$ into $x = 6 - 2y$:
   $x = 6 - 2(1)$
   $x = 6 - 2$
   $x = 4$.
   So, when $y=1$, $x=4$. The solution is the ordered pair (4, 1).
   Check: $4 + 2(1) = 4 + 2 = 6$. (True)
3. Choose $y = 2$:
   Substitute $y=2$ into $x = 6 - 2y$:
   $x = 6 - 2(2)$
   $x = 6 - 4$
   $x = 2$.
   So, when $y=2$, $x=2$. The solution is the ordered pair (2, 2).
   Check: $2 + 2(2) = 2 + 4 = 6$. (True)
4. Choose $y = 3$:
   Substitute $y=3$ into $x = 6 - 2y$:
   $x = 6 - 2(3)$
   $x = 6 - 6$
   $x = 0$.
   So, when $y=3$, $x=0$. The solution is the ordered pair (0, 3).
   Check: $0 + 2(3) = 0 + 6 = 6$. (True)

Summary of Solutions in a Table:

Chosen y	Calculated x ($x = 6 - 2y$)	Solution (x, y)
0	6	(6, 0)
1	4	(4, 1)
2	2	(2, 2)
3	0	(0, 3)

Therefore, four different solutions are (6, 0), (4, 1), (2, 2), and (0, 3). (Note: Many other solutions exist, for instance, choosing $x=0$ gives $0 + 2y = 6 \implies 2y=6 \implies y=3$, leading to the solution (0, 3) again. Choosing $x=1$ gives $1 + 2y = 6 \implies 2y=5 \implies y=2.5$, giving the solution (1, 2.5)).

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

We need to find two pairs (x, y) that satisfy this equation. A simple approach is to find the points where the line crosses the axes (the intercepts).

• Find the y-intercept (set $x = 0$):
  $4(0) + 3y = 12$
  $0 + 3y = 12$
  $3y = 12$
  $y = \frac{12}{3}$
  $y = 4$.
  So, one solution is (0, 4).
• Find the x-intercept (set $y = 0$):
  $4x + 3(0) = 12$
  $4x + 0 = 12$
  $4x = 12$
  $x = \frac{12}{4}$
  $x = 3$.
  So, another solution is (3, 0).

Table of Solutions for $4x + 3y = 12$:

Chosen Value	Calculated Value	Solution (x, y)
x = 0	y = 4	(0, 4)
y = 0	x = 3	(3, 0)

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

We need two pairs (x, y) that satisfy this equation. Let's try setting one variable to zero.

• Set $x = 0$:
  $2(0) + 5y = 0$
  $0 + 5y = 0$
  $5y = 0$
  $y = 0$.
  So, one solution is (0, 0). (This means the line passes through the origin).
• Set $y = 0$:
  $2x + 5(0) = 0$
  $2x + 0 = 0$
  $2x = 0$
  $x = 0$.
  This gives the same solution (0, 0). We need a *different* solution. Let's choose another value for x (or y).
• Choose a non-zero value, e.g., $x = 5$:
  $2(5) + 5y = 0$
  $10 + 5y = 0$
  $5y = -10$
  $y = -10 / 5$
  $y = -2$.
  So, another solution is (5, -2).
  (Alternatively, we could choose $y = 2$: $2x + 5(2) = 0 \implies 2x + 10 = 0 \implies 2x = -10 \implies x = -5$. Solution: $(-5, 2)$.)

Table of Solutions for $2x + 5y = 0$:

Chosen Value	Calculated Value	Solution (x, y)
x = 0	y = 0	(0, 0)
x = 5	y = -2	(5, -2)

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(iii) $3y + 4 = 0$

First, let's solve this equation for y:
$3y = -4$
$y = -\frac{4}{3}$
This equation states that the value of y *must* be $-\frac{4}{3}$. It doesn't depend on x at all. This means that for *any* value of x we choose, the value of y will always be $-\frac{4}{3}$.
We can write the equation in two variables as $0x + 3y + 4 = 0$.

• Choose $x = 0$:
  The equation requires $y = -\frac{4}{3}$.
  Solution 1: $(0, -\frac{4}{3})$.
• Choose $x = 3$: (We can choose any value for x)
  The equation still requires $y = -\frac{4}{3}$.
  Solution 2: $(3, -\frac{4}{3})$.

Table of Solutions for $3y + 4 = 0$ (or $y = -4/3$):

Chosen x	Required y	Solution (x, y)
0	$-\frac{4}{3}$	$(0, -\frac{4}{3})$
3	$-\frac{4}{3}$	$(3, -\frac{4}{3})$

The correct option is (iii) infinitely many solutions.

Why:

The given equation, y = 3x + 5, is a linear equation in two variables (x and y). A key property of such equations is that they represent a straight line on the Cartesian plane.

A solution to this equation is any pair of values (x, y) that makes the equation true. We can demonstrate that there are infinite such pairs:

• We can choose any real number for the value of x.
• For each chosen value of x, the equation provides a rule ($y = 3x + 5$) to calculate a unique corresponding value for y.
• Since there are infinitely many real numbers that we can choose for x, there will be infinitely many corresponding values of y.
• Each resulting (x, y) pair is a distinct solution to the equation.

Let's see a few examples:

• If we choose x = 0, then y = 3(0) + 5 = 0 + 5 = 5. So, (0, 5) is a solution.
• If we choose x = 1, then y = 3(1) + 5 = 3 + 5 = 8. So, (1, 8) is a solution.
• If we choose x = -1, then y = 3(-1) + 5 = -3 + 5 = 2. So, (-1, 2) is a solution.
• If we choose x = $\frac{1}{3}$, then y = 3($\frac{1}{3}$) + 5 = 1 + 5 = 6. So, ($\frac{1}{3}$, 6) is a solution.

This process can be continued indefinitely by choosing any real value for x (like fractions, decimals, irrational numbers), always yielding a corresponding y value and thus another solution. Therefore, the equation y = 3x + 5 has infinitely many solutions.

To find solutions for these linear equations in two variables, we can choose a value for one variable and solve for the other. We need to find four distinct pairs (x, y) for each equation.

(i) $2x + y = 7$

It's easiest to first isolate y:

$y = 7 - 2x$

Now, choose values for x and calculate y:

• Let $x = 0$: $y = 7 - 2(0) = 7 - 0 = 7$. Solution: (0, 7).
• Let $x = 1$: $y = 7 - 2(1) = 7 - 2 = 5$. Solution: (1, 5).
• Let $x = 2$: $y = 7 - 2(2) = 7 - 4 = 3$. Solution: (2, 3).
• Let $x = 3$: $y = 7 - 2(3) = 7 - 6 = 1$. Solution: (3, 1).

Summary Table for $2x + y = 7$:

Chosen x	Calculated y ($y = 7 - 2x$)	Solution (x, y)
0	7	(0, 7)
1	5	(1, 5)
2	3	(2, 3)
3	1	(3, 1)

Four solutions are (0, 7), (1, 5), (2, 3), (3, 1).

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(ii) $πx + y = 9$

Isolate y:

$y = 9 - πx$

Choose values for x and calculate y (solutions will involve π):

• Let $x = 0$: $y = 9 - π(0) = 9 - 0 = 9$. Solution: (0, 9).
• Let $x = 1$: $y = 9 - π(1) = 9 - π$. Solution: (1, 9 - π).
• Let $x = 2$: $y = 9 - π(2) = 9 - 2π$. Solution: (2, 9 - 2π).
• Let $x = \frac{9}{π}$: $y = 9 - π(\frac{9}{π}) = 9 - 9 = 0$. Solution: ($\frac{9}{π}$, 0).

Summary Table for $πx + y = 9$:

Chosen x	Calculated y ($y = 9 - πx$)	Solution (x, y)
0	9	(0, 9)
1	9 - π	(1, 9 - π)
2	9 - 2π	(2, 9 - 2π)
$\frac{9}{π}$	0	($\frac{9}{π}$, 0)

Four solutions are (0, 9), (1, 9 - π), (2, 9 - 2π), ($\frac{9}{π}$, 0).

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(iii) $x = 4y$

It's convenient to choose values for y and find the corresponding x.

• Let $y = 0$: $x = 4(0) = 0$. Solution: (0, 0).
• Let $y = 1$: $x = 4(1) = 4$. Solution: (4, 1).
• Let $y = -1$: $x = 4(-1) = -4$. Solution: (-4, -1).
• Let $y = 2$: $x = 4(2) = 8$. Solution: (8, 2).

Summary Table for $x = 4y$:

Chosen y	Calculated x ($x = 4y$)	Solution (x, y)
0	0	(0, 0)
1	4	(4, 1)
-1	-4	(-4, -1)
2	8	(8, 2)

Four solutions are (0, 0), (4, 1), (-4, -1), (8, 2).

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

(i) $(0, 2)$

Substitute $x=0, y=2$ into the LHS:

LHS = $0 - 2(2) = 0 - 4 = -4$.

Since LHS = $-4$ and RHS = $4$, $-4 \neq 4$.

Therefore, $(0, 2)$ is not a solution.

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(ii) $(2, 0)$

Substitute $x=2, y=0$ into the LHS:

LHS = $2 - 2(0) = 2 - 0 = 2$.

Since LHS = $2$ and RHS = $4$, $2 \neq 4$.

Therefore, $(2, 0)$ is not a solution.

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(iii) $(4, 0)$

Substitute $x=4, y=0$ into the LHS:

LHS = $4 - 2(0) = 4 - 0 = 4$.

Since LHS = $4$ and RHS = $4$, LHS = RHS.

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

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

Substitute $x=\sqrt{2}, y=4\sqrt{2}$ into the LHS:

LHS = $\sqrt{2} - 2(4\sqrt{2}) = \sqrt{2} - 8\sqrt{2} = (1-8)\sqrt{2} = -7\sqrt{2}$.

Since LHS = $-7\sqrt{2}$ and RHS = $4$, $-7\sqrt{2} \neq 4$.

Therefore, $(\sqrt{2}, 4\sqrt{2})$ is not a solution.

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(v) $(1, 1)$

Substitute $x=1, y=1$ into the LHS:

LHS = $1 - 2(1) = 1 - 2 = -1$.

Since LHS = $-1$ and RHS = $4$, $-1 \neq 4$.

Therefore, $(1, 1)$ is not a solution.

Given the equation $2x + 3y = k$.

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

This means that when we substitute $x = 2$ and $y = 1$ into the equation, the equality must hold true.

Substitute the values:

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

Calculate the left side:

$4 + 3 = k$

$7 = k$

Therefore, the value of k is 7.

Solution:

The given point is (1, 2). This means we have $x=1$ and $y=2$. We need to find linear equations in two variables, $x$ and $y$, for which this point is a solution.

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Finding Equations of Lines

We can create such equations by establishing a relationship between $x$ and $y$ that holds true for the given values. Here are a few examples:

Example 1: Sum of coordinates

Let's consider an equation of the form $x + y = c$.

Substituting the values $x=1$ and $y=2$:

$1 + 2 = 3$

So, one possible equation is $x + y = 3$.

Example 2: A multiple relationship

Let's consider an equation of the form $y = mx$.

Substituting the values $x=1$ and $y=2$:

$2 = m(1) \implies m = 2$

So, another possible equation is $y = 2x$.

Example 3: A different linear combination

Let's consider an equation of the form $3x - y = c$.

Substituting the values $x=1$ and $y=2$:

$3(1) - 2 = 3 - 2 = 1$

So, another possible equation is $3x - y = 1$.

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How many such equations are there?

There are an infinite number of such equations.

Reason:

Geometrically, an infinite number of straight lines can be drawn passing through a single point. Each of these lines has a unique equation.

Algebraically, we can see this by considering the general form of a linear equation, $ax + by = c$. For the point (1, 2) to lie on this line, the equation must be satisfied:

$a(1) + b(2) = c$

$a + 2b = c$

We can choose any values we want for 'a' and 'b' (as long as they are not both zero), and this equation will give us a corresponding value for 'c'. Since there are infinite choices for 'a' and 'b', we can generate an infinite number of unique linear equations.

Solution:

The given equation is $x + y = 7$. This is a linear equation in two variables, and its graph will be a straight line. To draw the graph, we need to find at least two points (or solutions) that satisfy this equation. Finding a third point is a good way to verify that the line is drawn correctly.

A solution is an ordered pair $(x, y)$ that makes the equation true. We can find solutions by choosing a value for one variable and then calculating the corresponding value for the other variable. It is often helpful to express $y$ in terms of $x$ first:

$y = 7 - x$

Now, let's find some solutions:

• If we choose $x = 0$, then $y = 7 - 0 = 7$. So, one point is (0, 7).
• If we choose $y = 0$, then $x + 0 = 7$, which means $x = 7$. So, another point is (7, 0).
• If we choose $x = 3$, then $y = 7 - 3 = 4$. So, a third point is (3, 4).

We can summarize these solutions in a table:

x	y ($y = 7-x$)	Point (x, y)
0	7	(0, 7)
7	0	(7, 0)
3	4	(3, 4)

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Now, we plot these points on a Cartesian plane and draw a straight line that passes through them. This line represents the graph of the equation $x + y = 7$. Every point on this line is a solution to the equation.

Solution:

This problem describes a fundamental concept in physics, which we can express mathematically and represent graphically.

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Step 1: Writing the Equation

The problem states that the force applied on a body is directly proportional to the acceleration produced in the body.

Let's represent the force by y and the acceleration by x.

The statement "y is directly proportional to x" can be written mathematically as:

$y \propto x$

To turn this proportionality into an equation, we introduce a constant of proportionality, which we will call k. This constant is a fixed value that depends on the properties of the body (in this case, its mass).

Since the value of the constant (mass) is not given, we can assume a value for it to draw the graph. Let's take the constant of proportionality k = 3 units (e.g., a mass of 3 kg).

So, the specific equation we will plot is:

$y = 3x$

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Step 2: Finding Solutions for the Equation

To draw the graph of this linear equation, we need to find at least two points (solutions) that satisfy it. We can create a table of values by choosing different values for x (acceleration) and calculating the corresponding values for y (force).

Acceleration x (in units)	Force y ($y=3x$) (in units)	Point (x, y)
0	$3 \times 0 = 0$	(0, 0)
1	$3 \times 1 = 3$	(1, 3)
2	$3 \times 2 = 6$	(2, 6)
-1	$3 \times (-1) = -3$	(-1, -3)

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Step 3: Plotting the Graph

Now, we will plot the points (0, 0), (1, 3), (2, 6), and (-1, -3) on a Cartesian plane. We will represent acceleration (x) on the horizontal axis and force (y) on the vertical axis. After plotting the points, we join them with a straight line.

The resulting graph is a straight line that passes through the origin. This is the characteristic graph for a direct proportionality relationship. The slope of the line (which is 3 in this case) represents the constant of proportionality (the mass of the body).

To find the correct equation for each graph, we need to check which equation is satisfied by the points shown on the graph's line.

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(a) For Graph (i)

The points marked on the line are (1, 2), (0, 0), and (-1, -2).

Let's test the given equations with one of the points, say (1, 2) where x=1 and y=2.

• (i) $x + y = 0 \implies 1 + 2 = 3 \neq 0$. (Incorrect)
• (ii) $y = 2x \implies 2 = 2(1) \implies 2 = 2$. (Correct)
• (iii) $y = x \implies 2 = 1$. (Incorrect)
• (iv) $y = 2x + 1 \implies 2 = 2(1) + 1 \implies 2 = 3$. (Incorrect)

Let's verify with another point, (-1, -2), for the equation $y=2x$.

$y = 2x \implies -2 = 2(-1) \implies -2 = -2$. (Correct)

The correct equation for graph (i) is (ii) $y = 2x$.

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(b) For Graph (ii)

The points marked on the line are (1, 6), (0, 4), and (-2, 0).

Let's test the given equations with the point (0, 4) where x=0 and y=4.

• (i) $x + y = 0 \implies 0 + 4 = 4 \neq 0$. (Incorrect)
• (ii) $y = 2x \implies 4 = 2(0) \implies 4 = 0$. (Incorrect)
• (iii) $y = 2x + 4 \implies 4 = 2(0) + 4 \implies 4 = 4$. (Correct)
• (iv) $y = x - 4 \implies 4 = 0 - 4 \implies 4 = -4$. (Incorrect)

Let's verify with another point, (-2, 0), for the equation $y = 2x + 4$.

$y = 2x + 4 \implies 0 = 2(-2) + 4 \implies 0 = -4 + 4 \implies 0 = 0$. (Correct)

The correct equation for graph (ii) is (iii) $y = 2x + 4$.

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(c) For Graph (iii)

The points marked on the line are (2, 0), (1, -2), (0, -4), and (-1, -6).

Let's test the given equations with the point (2, 0) where x=2 and y=0.

• (i) $x + y = 0 \implies 2 + 0 = 2 \neq 0$. (Incorrect)
• (ii) $y = 2x \implies 0 = 2(2) \implies 0 = 4$. (Incorrect)
• (iii) $y = 2x + 1 \implies 0 = 2(2) + 1 \implies 0 = 5$. (Incorrect)
• (iv) $y = 2x - 4 \implies 0 = 2(2) - 4 \implies 0 = 4 - 4 \ $$ \implies 0 = 0$. (Correct)

Let's verify with another point, (0, -4), for the equation $y = 2x - 4$.

$y = 2x - 4 \implies -4 = 2(0) - 4 \implies -4 = -4$. (Correct)

The correct equation for graph (iii) is (iv) $y = 2x - 4$.

To draw the graph of a linear equation, we need to find at least two coordinate points that satisfy the equation. We can then plot these points and draw a straight line through them.

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(i) $x + y = 4$

We can write this as $y = 4 - x$. Let's find some points:

• If $x=0$, $y=4-0=4$. Point: (0, 4)
• If $y=0$, $x=4$. Point: (4, 0)
• If $x=2$, $y=4-2=2$. Point: (2, 2)

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(ii) $x - y = 2$

We can write this as $y = x - 2$. Let's find some points:

• If $x=0$, $y=0-2=-2$. Point: (0, -2)
• If $y=0$, $x=2$. Point: (2, 0)
• If $x=4$, $y=4-2=2$. Point: (4, 2)

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(iii) $y = 3x$

Let's find some points:

• If $x=0$, $y=3(0)=0$. Point: (0, 0)
• If $x=1$, $y=3(1)=3$. Point: (1, 3)
• If $x=-1$, $y=3(-1)=-3$. Point: (-1, -3)

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(iv) $3 = 2x + y$

We can write this as $y = 3 - 2x$. Let's find some points:

• If $x=0$, $y=3-2(0)=3$. Point: (0, 3)
• If $y=0$, $3=2x \implies x=1.5$. Point: (1.5, 0)
• If $x=2$, $y=3-2(2)=3-4=-1$. Point: (2, -1)

Part 1: Give the equations of two lines passing through (2, 14).

To find the equation of a line passing through the point (2, 14), we need to find a linear relationship between $x$ and $y$ that is true when $x=2$ and $y=14$. We can create an infinite number of such equations. Here are two examples:

Equation 1:

Let's form an equation based on the sum of the coordinates.

$x + y = c$

Substitute the point (2, 14) into the equation to find the value of c:

$2 + 14 = 16$

So, one possible equation is $x + y = 16$.

Equation 2:

Let's form an equation where y is a multiple of x.

$y = kx$

Substitute the point (2, 14) into the equation to find the value of k:

$14 = k(2)$

$k = \frac{14}{2} = 7$

So, another possible equation is $y = 7x$ (or $7x - y = 0$).

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Part 2: How many more such lines are there, and why?

There are an infinite number of such lines.

Reason:

Geometrically, an infinite number of unique straight lines can be drawn passing through a single, fixed point. Imagine a point on a piece of paper; you can pivot a ruler around that point in countless different directions, and each position represents a different line.

Algebraically, we can represent a line with the general equation $ax + by = c$. For the line to pass through (2, 14), the equation $a(2) + b(14) = c$ must be true. We can choose an infinite number of different values for 'a' and 'b' (as long as they are not both zero), and each pair will give a corresponding value of 'c', creating a new, unique equation. For example, if we choose $a=3$ and $b=1$, then $c = 3(2) + 1(14) = 20$, giving the equation $3x+y=20$. Since there are infinite choices for 'a' and 'b', there are infinite possible equations.

Given:

The point (3, 4) lies on the graph of the equation $3y = ax + 7$.

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To Find:

The value of 'a'.

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Solution:

If a point lies on the graph of an equation, it means that the coordinates of that point are a solution to the equation. In this case, the point is (3, 4), which means we have:

$x = 3$

$y = 4$

We can substitute these values of $x$ and $y$ into the given equation to find the value of 'a'.

The equation is:

$3y = ax + 7$

Substitute $x=3$ and $y=4$ into the equation:

$3(4) = a(3) + 7$

Now, we simplify and solve for 'a':

$12 = 3a + 7$

Subtract 7 from both sides of the equation:

$12 - 7 = 3a$

$5 = 3a$

Divide by 3 to isolate 'a':

$a = \frac{5}{3}$

Therefore, the value of 'a' is $\frac{5}{3}$.

Solution:

Part 1: Writing the Linear Equation

Let's break down the taxi fare structure:

Total distance covered = $x$ km

Total fare = $\textsf{₹} y$

The fare for the first kilometre is fixed at $\textsf{₹} 8$.

The remaining distance after the first kilometre is $(x - 1)$ km.

The fare for this subsequent distance is charged at a rate of $\textsf{₹} 5$ per km. So, the cost for the subsequent distance is $5 \times (x - 1)$.

The total fare (y) is the sum of the fare for the first kilometre and the fare for the subsequent distance.

Total Fare = (Fare for first km) + (Fare for subsequent distance)

$y = 8 + 5(x - 1)$

Now, let's simplify this equation:

$y = 8 + 5x - 5$

$y = 5x + 3$

This is the linear equation that represents the given information.

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Part 2: Drawing the Graph

To draw the graph of the equation $y = 5x + 3$, we need to find at least two points that satisfy it. Let's find a few points by choosing different values for $x$ (distance).

• If $x = 0$ (no distance covered), $y = 5(0) + 3 = 3$. Point: (0, 3). (This could be considered a base meter charge).
• If $x = 1$ (1 km covered), $y = 5(1) + 3 = 8$. Point: (1, 8).
• If $x = 2$ (2 km covered), $y = 5(2) + 3 = 13$. Point: (2, 13).

We can summarize these points in a table:

Distance x (in km)	Fare y (in $\textsf{₹}$)	Point (x, y)
0	3	(0, 3)
1	8	(1, 8)
2	13	(2, 13)

Now, we will plot these points on a graph. The x-axis will represent the distance in km, and the y-axis will represent the fare in $\textsf{₹}$. Since distance and fare cannot be negative, we will only use the first quadrant.

The graph is a straight line starting from the point (0, 3) and rising, which represents the linear relationship between the distance covered and the total taxi fare.

To determine the correct equation for each graph, we need to check which of the given equations is satisfied by the points plotted on the line.

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For Fig. 4.6

The graph shows a line passing through the points (1, -1), (0, 0), and (-1, 1).

Let's test these points against the given equations. We can use the point (1, -1), where $x=1$ and $y=-1$.

• (i) $y = x$: $-1 = 1$. (Incorrect)
• (ii) $x + y = 0$: $1 + (-1) = 0 \implies 0 = 0$. (Correct)
• (iii) $y = 2x$: $-1 = 2(1) \implies -1 = 2$. (Incorrect)
• (iv) $2 + 3y = 7x$: $2 + 3(-1) = 7(1) \implies 2 - 3 = 7 \implies -1 = 7$. (Incorrect)

Let's verify with another point, (-1, 1), for the equation $x+y=0$:

$-1 + 1 = 0$. This is also correct.

The correct equation for the graph in Fig. 4.6 is (ii) $x + y = 0$.

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For Fig. 4.7

The graph shows a line passing through the points (2, 0), (0, 2), and (-1, 3).

Let's test these points against the given equations. We can use the point (0, 2), where $x=0$ and $y=2$.

• (i) $y = x + 2$: $2 = 0 + 2 \implies 2 = 2$. (Correct)
• (ii) $y = x – 2$: $2 = 0 - 2 \implies 2 = -2$. (Incorrect)
• (iii) $y = –x + 2$: $2 = -0 + 2 \implies 2 = 2$. (Correct)
• (iv) $x + 2y = 6$: $0 + 2(2) = 6 \implies 4 = 6$. (Incorrect)

We have two possible correct equations: (i) and (iii). Let's test them with another point, (2, 0), where $x=2$ and $y=0$.

• For equation (i) $y = x + 2$: $0 = 2 + 2 \implies 0 = 4$. (Incorrect)
• For equation (iii) $y = -x + 2$: $0 = -(2) + 2 \implies 0 = 0$. (Correct)

The equation that satisfies all the points is $y = -x + 2$.

The correct equation for the graph in Fig. 4.7 is (iii) $y = –x + 2$.

Solution:

Part 1: Expressing the information as a linear equation

We are given that the work done by a body is directly proportional to the distance travelled.

Let the work done be represented by y.

Let the distance travelled be represented by x.

The relationship "work is directly proportional to distance" can be written as:

$y \propto x$

To convert this into an equation, we introduce a constant of proportionality. The problem states that this constant is the force applied, which is given as 5 units.

So, the equation is:

$y = 5x$

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Part 2: Drawing the graph

To draw the graph of the linear equation $y = 5x$, we need to find at least two points that satisfy it. Since distance (x) and work done (y) are physical quantities, they cannot be negative. Therefore, we will only consider non-negative values for x.

• If $x = 0$ (distance is 0 units), then $y = 5(0) = 0$. Point: (0, 0).
• If $x = 1$ (distance is 1 unit), then $y = 5(1) = 5$. Point: (1, 5).
• If $x = 2$ (distance is 2 units), then $y = 5(2) = 10$. Point: (2, 10).

Distance x (units)	Work Done y (units)	Point (x, y)
0	0	(0, 0)
1	5	(1, 5)
2	10	(2, 10)

Now, we plot these points on a graph where the x-axis represents the distance and the y-axis represents the work done. Then we draw a line starting from the origin and passing through these points.

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Part 3: Reading values from the graph

(i) Work done when the distance travelled is 2 units:

To find this from the graph, we locate 2 on the x-axis (Distance). We move vertically up from this point until we intersect the graph line. Then, we move horizontally from the intersection point to the y-axis (Work Done). The value we read on the y-axis is 10.

So, the work done is 10 units.

(ii) Work done when the distance travelled is 0 unit:

We locate 0 on the x-axis, which is the origin. The corresponding point on the graph is also at the origin. The y-value at this point is 0.

So, the work done is 0 units.

Solution:

Part 1: Writing the Linear Equation

We are given that Yamini and Fatima together contributed $\textsf{₹}$100.

Let Yamini's contribution be $\textsf{₹} x$.

Let Fatima's contribution be $\textsf{₹} y$.

According to the problem statement, the sum of their contributions is 100. We can express this as a linear equation:

$x + y = 100$

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Part 2: Drawing the Graph

To draw the graph of the linear equation $x + y = 100$, we need to find at least two pairs of values for $x$ and $y$ that satisfy the equation. Since the contributions cannot be negative, we will only consider non-negative values for $x$ and $y$.

We can rewrite the equation as $y = 100 - x$.

• If Yamini contributes nothing ($x=0$), then Fatima contributes $y = 100 - 0 = 100$. So, one point is (0, 100).
• If Fatima contributes nothing ($y=0$), then Yamini contributes $x = 100$. So, another point is (100, 0).
• If Yamini contributes $\textsf{₹}$50 ($x=50$), then Fatima contributes $y = 100 - 50 = 50$. So, a third point is (50, 50).

We can summarize these solutions in a table:

Yamini's Contribution (x in $\textsf{₹}$)	Fatima's Contribution (y in $\textsf{₹}$)	Point (x, y)
0	100	(0, 100)
100	0	(100, 0)
50	50	(50, 50)

Now, we will plot these points on a graph. The x-axis will represent Yamini's contribution and the y-axis will represent Fatima's contribution. We will choose a suitable scale, for example, 1 unit = $\textsf{₹}$ 20 on both axes.

The graph is a line segment connecting the points (0, 100) and (100, 0). Every point on this line segment represents a possible combination of contributions from Yamini and Fatima that adds up to $\textsf{₹}$100.

The given linear equation is $F = \left(\frac{9}{5}\right)C + 32$, where C is the temperature in Celsius and F is the temperature in Fahrenheit.

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(i) Draw the graph of the linear equation.

To draw the graph, we will use Celsius (C) for the x-axis and Fahrenheit (F) for the y-axis. Let's find some points that satisfy the equation $y = \frac{9}{5}x + 32$.

• If C = 0, then $F = \frac{9}{5}(0) + 32 = 32$. Point: (0, 32).
• If C = -10, then $F = \frac{9}{5}(-10) + 32 = -18 + 32 = 14$. Point: (-10, 14).
• If C = 10, then $F = \frac{9}{5}(10) + 32 = 18 + 32 = 50$. Point: (10, 50).

Celsius C (x)	Fahrenheit F (y)	Point (C, F)
0	32	(0, 32)
-10	14	(-10, 14)
10	50	(10, 50)

Now, we plot these points on a graph and draw a straight line through them.

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(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?

We substitute C = 30 into the equation:

$F = \frac{9}{5}(30) + 32 = 9 \times 6 + 32 = 54 + 32 = 86$.

The temperature is 86°F.

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(iii) If the temperature is 95°F, what is the temperature in Celsius?

We substitute F = 95 into the equation and solve for C:

$95 = \frac{9}{5}C + 32$

$95 - 32 = \frac{9}{5}C$

$63 = \frac{9}{5}C$

$C = \frac{63 \times 5}{9} = 7 \times 5 = 35$.

The temperature is 35°C.

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(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

• If C = 0°C, then $F = \frac{9}{5}(0) + 32 = 32$. The temperature is 32°F.
• If F = 0°F, then $0 = \frac{9}{5}C + 32 \implies -32 = \frac{9}{5}C \ $$ \implies C = \frac{-32 \times 5}{9} = -\frac{160}{9} \approx -17.8$. The temperature is approximately -17.8°C.

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(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Yes. To find this temperature, we need to find a value where C = F. Let's set C = F = x in the equation.

$x = \frac{9}{5}x + 32$

$x - \frac{9}{5}x = 32$

$\frac{5x - 9x}{5} = 32$

$\frac{-4x}{5} = 32$

$x = \frac{32 \times 5}{-4} = -8 \times 5 = -40$.

Yes, the temperature that is numerically the same is -40 degrees (-40°C = -40°F).

Solution:

First, we need to solve the given linear equation in one variable.

Solving the equation:

$2x + 1 = x - 3$

To solve for $x$, we gather the terms involving $x$ on one side and the constant terms on the other side.

Subtract $x$ from both sides:

$2x - x + 1 = -3$

$x + 1 = -3$

Subtract 1 from both sides:

$x = -3 - 1$

$x = -4$

The solution to the equation is $x = -4$.

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(i) Representation on the number line

The equation $x = -4$ is an equation in one variable. Its solution is a single point on the number line. We draw a number line and mark a distinct point at the position corresponding to -4.

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(ii) Representation on the Cartesian plane

To represent the solution on the Cartesian plane (a two-dimensional plane with x and y axes), we need to think of the equation $x = -4$ as a linear equation in two variables. We can write it as:

$x + 0y = -4$

This equation tells us that for any value of $y$, the value of $x$ must always be -4. Let's find some points that satisfy this equation:

• If $y = 0$, $x = -4$. Point: (-4, 0)
• If $y = 2$, $x = -4$. Point: (-4, 2)
• If $y = -3$, $x = -4$. Point: (-4, -3)

When we plot these points on the Cartesian plane and join them, we get a straight line.

The graph of the equation $x = -4$ on the Cartesian plane is a vertical line parallel to the y-axis, passing through the point (-4, 0).

Solution:

The given equation is $y = 3$.

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(i) Geometric representation in one variable

When we consider $y = 3$ as an equation in one variable, we are thinking of a number line for the variable 'y'. The solution is a single, specific point on this number line.

The geometric representation is a point on the number line located at the position corresponding to the number 3.

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(ii) Geometric representation in two variables

To represent $y = 3$ as an equation in two variables (x and y), we can write it in the standard form $ax + by = c$. This would be:

$0x + y = 3$

This equation tells us that for any value of the variable $x$, the value of $y$ must always be 3. Let's find some coordinate points that are solutions to this equation:

• If $x = 0$, $y = 3$. Point: (0, 3)
• If $x = 2$, $y = 3$. Point: (2, 3)
• If $x = -4$, $y = 3$. Point: (-4, 3)

When we plot these points on a Cartesian plane and join them, we get a straight line.

The geometric representation of $y = 3$ in two variables is a horizontal line parallel to the x-axis, at a distance of 3 units above it.

Solution:

First, let's solve the given equation for x.

$2x + 9 = 0$

$2x = -9$

$x = -\frac{9}{2} = -4.5$

The equation simplifies to $x = -4.5$.

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(i) Geometric representation in one variable

When we consider $x = -4.5$ as an equation in one variable, we represent it on a number line for the variable 'x'. The solution is a single point on this line.

The geometric representation is a point on the number line located at the position -4.5, which is exactly halfway between -4 and -5.

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(ii) Geometric representation in two variables

To represent $x = -4.5$ as an equation in two variables (x and y), we can write it in the form:

$x + 0y = -4.5$

This equation indicates that for any value of the variable $y$, the value of $x$ must always be -4.5. Let's find some coordinate points that satisfy this equation:

• If $y = 0$, $x = -4.5$. Point: (-4.5, 0)
• If $y = 3$, $x = -4.5$. Point: (-4.5, 3)
• If $y = -2$, $x = -4.5$. Point: (-4.5, -2)

When we plot these points on a Cartesian plane and join them, we get a straight line.

The geometric representation of $2x + 9 = 0$ (or $x = -4.5$) in two variables is a vertical line parallel to the y-axis, at a distance of 4.5 units to the left of it.