PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Exercise 3.1

Question 1

Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:

Let the number of boys be x.

Let the number of girls be y.

Step 1: Form equations according to the question:

1. Total number of students is 10: x + y = 10 … (1)

2. The no. of girls 4 more than the no. of boys: y = x + 4 … (2)

Step 2: Solve graphically

To solve these equations graphically, follow these steps:

For equation (1): x + y = 10

Find two points:

If x = 0, then y = 10 → (0, 10)

If x = 6, then y = 4 → (6, 4)

For equation (2): y = x + 4

Find two points:

If x = 0, then y = 4 → (0, 4)

If x = 2, then y = 6 → (2, 6)

Now, draw both lines on a graph paper:

Plot the points (0,10) and (6,4) and draw line for equation (1)

Plot the points (0,4) and (2,6) and draw line for equation (2)

The point where the two lines intersect is the solution.

Step 3: Read the solution

The two lines intersect at point (3, 7)

Answer: Number of boys = 3; Number of girls = 7

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Q1 (ii)

Form the pair of linear equations in the following problems, and find their solutions graphically.

5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.

Solution:

Let the cost of one pencil be x rupees.

Let the cost of one pen be y rupees.

Step 1: Form equations According to the question:

1. 5 pencils and 7 pens cost ₹50:

5x + 7y = 50 … (1)

2. 7 pencils and 5 pens cost ₹46:

7x + 5y = 46 … (2)

Step 2: Solve graphically

To solve these equations graphically, we will find two points for each line.

For equation (1): 5x + 7y = 50

Choose values for x and solve for y:

If x = 1 → 5(1) + 7y = 50 → 7y = 45 → y = 6.43 (approx)

If x = 5 → 5(5) + 7y = 50 → 25 + 7y = 50 → 7y = 25 → y = 3.57 (approx)

Points: (1, 6.43), (5, 3.57)

For equation (2): 7x + 5y = 46

Choose values for x and solve for y:

If x = 4 → 7(4) + 5y = 46 → 28 + 5y = 46 → 5y = 18 → y = 3.6

If x = 5 → 7(5) + 5y = 46 → 35 + 5y = 46 → 5y = 11 → y = 2.2

Points: (4, 3.6), (5, 2.2)

Step 3: Draw the graph

Plot the points for each equation on graph paper.

Draw lines through the points for both equations.

The point where the two lines intersect gives the solution.

Step 4: Read the solution

The lines intersect at the point (3, 5)

Answer: Cost of one pencil = ₹3; Cost of one pen = ₹5

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Question 2:

On comparing the ratios find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) Solve equations: 5x - 4y + 8 = 0 and 7x + 6y - 9 = 0

Solution:

First, write down the coefficients:

From equation 1:

a1 = 5, b1 = -4, c1 = 8

From equation 2:

a2 = 7, b2 = 6, c2 = -9

Now compare the ratios:

a1/a2 = 5/7

b1/b2 = -4/6 = -2/3

c1/c2 = 8/-9

Now check the relationship:

Since a1/a2 is not equal to b1/b2, the lines intersect at one point.

Answer: The lines intersect at a point.

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Q2

(ii) Equations: 9x + 3y + 12 = 0 and 18x + 6y + 24 = 0

Solution:

First, write down the coefficients:

From equation 1: a1 = 9, b1 = 3, c1 = 12

From equation 2: a2 = 18, b2 = 6, c2 = 24

Now compare the ratios:

a1\a2 = 9\18 = 1\2

b1\b2 = 3\6 = 1\2

c1\c2 = 12\24 = 1\2

Now check the relationship:

Since all three ratios are equal, the lines are coincident.

Answer: The lines are coincident.

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Q2

(iii) Equations: 6x - 3y + 10 = 0 and 2x - y + 9 = 0

Solution:

First, write down the coefficients:

From equation 1:

a1 = 6, b1 = -3, c1 = 10

From equation 2:

a2 = 2, b2 = -1, c2 = 9

Now compare the ratios:

a1\a2 = 6\2 = 3\1

b1\b2 = -3\-1 = 3\1

c1\c2 = 10\9

Now check the relationship:

a1\a2 equals b1\b2, but not equal to c1\c2

Answer: The lines are parallel.

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Q3:

On comparing the ratios find out whether the following pair of linear equations are consistent or inconsistent:

(i) 3x + 2y = 5; 2x – 3y = 7

Solution:

First, write both equations in standard form:

Equation 1: 3x + 2y − 5 = 0

Equation 2: 2x − 3y − 7 = 0

Now, identify the coefficients:

From equation 1:

a1 = 3, b1 = 2, c1 = -5

From equation 2:

a2 = 2, b2 = -3, c2 = -7

Now compare the ratios:

a1/a2 = 3/2

b1/b2 = 2/3

c1/c2 = -5/7 = 5/7

Since a1/a2 is not equal to b1/b2, the lines intersect at a point.

Answer: The pair of linear equations is consistent.

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Q3:

(ii) 2x – 3y = 8; 4x – 6y = 9

Solution:

First, write both equations in standard form:

Equation 1: 2x − 3y − 8 = 0

Equation 2: 4x − 6y − 9 = 0

Now, identify the coefficients:

From equation 1:

a1 = 2, b1 = -3, c1 = -8

From equation 2:

a2 = 4, b2 = -6, c2 = -9

Now compare the ratios:

a1/ a2 = 2/4 = 1/2

b1/ b2 = -3/-6 = 1/2

c1/c2 = -8/-9

Since a1/a2 is equal to b1/b2, but not equal to c1/c2, the lines are parallel.

Answer: The pair of linear equations is inconsistent.

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Q3:

(iii) ; 9x – 10y = 14

Solution:

First, write both equations in standard form:

Equation 1: (3/2)x + (5/3)y − 7 = 0
Equation 2: 9x − 10y − 14 = 0

Now, identify the coefficients:

From equation 1:
a1 = 3/2, b1 = 5/3, c1 = -7

From equation 2:
a2 = 9, b2 = -10, c2 = -14

Now compare the ratios:

a1/a2 = (3/2)/9 = 3/18 = 1/6
b1/b2 = (5/3)/-10 = 5/-30 = -1/6
c1/c2 = -7/-14 = 1/2

Since a1/a2 is not equal to b1/b2, the lines intersect at a point.

Answer: The pair of linear equations is consistent.

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(iv) 5x – 3y = 11 ; – 10x + 6y = –22

Solution:

First, write both equations in standard form:

Equation 1: 5x − 3y − 11 = 0

Equation 2: −10x + 6y + 22 = 0

Now, identify the coefficients:

From equation 1:

a1 = 5, b1 = -3, c1 = -11

From equation 2:

a2 = -10, b2 = 6, c2 = 22

Now compare the ratios:

a1\a2 = 5\-10 = -1\ 2

b1\b2 = -3\6 = -1\ 2

c1\c2 = -11\22 = -1\2

Since all three ratios are equal, the lines are coincident.

Answer: The pair of linear equations is consistent and the lines are coincident.

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Q3

(v) (4/3)x + 2y = 8; 2x + 3y = 12

Solution:

We compare these equations with the general form:

a₁x + b₁y = c₁

a₂x + b₂y = c₂

So, we identify:

a₁ = 4/3, b₁ = 2, c₁ = 8

a₂ = 2, b₂ = 3, c₂ = 12

Step 1: Compare the ratios

a₁/a₂ = (4/3) ÷ 2 = (4/3) × (1/2) = 2/3

b₁/b₂ = 2 ÷ 3 = 2/3

c₁/c₂ = 8 ÷ 12 = 2/3

Step 2: Analyze the results

Since all three ratios are equal (2/3), i.e.,

a₁/a₂ = b₁/b₂ = c₁/c₂

Answer: This means the pair of equations is consistent.

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Question 4:

(i) x + y = 5 ; 2x + 2y = 10

Solution

Step 1: Write in standard form

These are already in standard form:

Equation 1: a₁ = 1, b₁ = 1, c₁ = 5

Equation 2: a₂ = 2, b₂ = 2, c₂ = 10

Step 2: Compare the ratios

a₁/a₂ = 1 / 2 = 0.5

b₁/b₂ = 1 / 2 = 0.5

c₁/c₂ = 5 / 10 = 0.5

All three ratios are equal:

a₁/a₂ = b₁/b₂ = c₁/c₂ = 0.5

Step 3: Analyze the result

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the two equations represent the same line.

Therefore, the system is consistent and dependent.

This means there are infinitely many solutions, and every solution of the first equation is also a solution of the second.

Step 4: Graphical solution

To graph:

Equation 1: x + y = 5

→ When x = 0, y = 5 → point (0, 5)

→ When y = 0, x = 5 → point (5, 0)

Equation 2 is just a multiple of equation 1 (both represent the same line), so the graph will show both lines overlapping.

Conclusion:

The system is consistent and dependent.

It has infinitely many solutions.

Graphically, the two lines coincide (they are the same line).

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Q4

(ii) x – y = 8 ; 3x – 3y = 16

Solution:

Step 1: Write in standard form

Equation 1: a₁ = 1, b₁ = -1, c₁ = 8

Equation 2: a₂ = 3, b₂ = -3, c₂ = 16

Step 2: Compare the ratios

a₁/a₂ = 1 / 3 ≈ 0.33

b₁/b₂ = -1 / -3 = 1 / 3 ≈ 0.33

c₁/c₂ = 8 / 16 = 0.5

Now compare:

a₁/a₂ = b₁/b₂ (both are 1/3)

But c₁/c₂ ≠ a₁/a₂

Step 3: Analyze the result

Since a₁/a₂ = b₁/b₂ but c₁/c₂ is not equal, the equations represent parallel lines that never intersect.

Therefore, the system is inconsistent. There is no solution.

Graphically, the lines are parallel and do not meet.

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Q4

(iii) 2x + y – 6 = 0 ; 4x – 2y – 4 = 0

Solution

Step 1: Rewrite in standard form

Move constants to the right-hand side:

Equation 1: 2x + y = 6

Equation 2: 4x – 2y = 4

Now identify coefficients:

a₁ = 2, b₁ = 1, c₁ = 6

a₂ = 4, b₂ = -2, c₂ = 4

Step 2: Compare the ratios

a₁/a₂ = 2 / 4 = 1/2

b₁/b₂ = 1 / (-2) = -1/2

c₁/c₂ = 6 / 4 = 3/2

Step 3: Analyze the result

a₁/a₂ ≠ b₁/b₂

This means the lines are not parallel and they will intersect at a point.

So, the system is consistent and independent → it has a unique solution.

Step 4: Solve the system

We’ll solve by substitution or elimination. Let’s use substitution:

From equation 1: 2x + y = 6 → y = 6 – 2x

Substitute into equation 2:

4x – 2(6 – 2x) = 4

4x – 12 + 4x = 4

8x – 12 = 4

8x = 16

x = 2

Now substitute x = 2 into y = 6 – 2x:

y = 6 – 2(2) = 6 – 4 = 2

Answer:

The system is consistent and independent.

It has a unique solution: x = 2, y = 2

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Q4

(iv) 2x – 2y – 2 = 0 ; 4x – 4y – 5 = 0

Solution:

Step 1: Rewrite in standard form

Move constants to the right-hand side:

Equation 1: 2x – 2y = 2

Equation 2: 4x – 4y = 5

Now identify coefficients:

a₁ = 2, b₁ = –2, c₁ = 2

a₂ = 4, b₂ = –4, c₂ = 5

Step 2: Compare the ratios

a₁/a₂ = 2 / 4 = 1/2

b₁/b₂ = –2 / –4 = 1/2

c₁/c₂ = 2 / 5

Now compare:

a₁/a₂ = b₁/b₂ = 1/2

But c₁/c₂ = 2/5 ≠ 1/2

Step 3: Analyze the result

Since a₁/a₂ = b₁/b₂ but c₁/c₂ ≠ a₁/a₂, the lines are parallel and will never meet.

Therefore, the system is inconsistent.

Conclusion:

The system is inconsistent.

There is no solution.

Graphically, the lines are parallel and do not intersect.

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Question 5:

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution

Step 1: Let the variables represent the dimensions

Let the width of the rectangular garden be x meters.

Then the length is x + 4 meters (since it is 4 meters more than the width).

Step 2: Use the formula for perimeter

The perimeter P of a rectangle is: P = 2 × (length + width)

We're told that half the perimeter is 36 meters, so: (1/2) × P = 36

Multiply both sides by 2: P = 72

Now plug into the formula:

2 × (x + x + 4) = 72

2 × (2x + 4) = 72

4x + 8 = 72

4x = 72 – 8

4x = 64

x = 64 ÷ 4

x = 16

Step 3: Find the length

Length = x + 4 = 16 + 4 = 20

Answer: These are the dimensions of the garden.

Width = 16 meters

Length = 20 meters

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Question 6:

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

Solution:

Step 1: Rewrite the given equation in slope-intercept form (y = mx + c):

2x + 3y – 8 = 0

3y = -2x + 8

y = (-2/3)x + 8/3

So, the slope of the given line is -2/3.

Step 2: To get intersecting lines, choose another line with a different slope.

Let's take this equation:

3x + 2y – 7 = 0

Convert it to slope-intercept form:

3x + 2y – 7 = 0

2y = -3x + 7

y = (-3/2)x + 7/2

This line has a slope of -3/2.

Step 3: Since -2/3 and -3/2 are different slopes, the two lines are not parallel.

Therefore, the lines will intersect at one point.

Answer: Another linear equation is: 3x + 2y – 7 = 0

This equation, together with the given one, represents intersecting lines.

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Q6:

(ii) Parallel lines

Given: 2x + 3y – 8 = 0

Write another linear equation so the pair forms parallel lines.

Solution:

1. The slope of the given line is –2/3 (from 2x + 3y – 8 = 0).

2. For parallel lines, the new equation must have the same slope.

3. Keep the same left-hand side: 2x + 3y, but change the constant.

4. New equation: 2x + 3y – 12 = 0

Answer: 2x + 3y – 12 = 0

This line is parallel to the given line.

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Q6:

(iii) coincident lines

Given the linear equation: 2x + 3y – 8 = 0

Solution:

Another linear equation that represents coincident lines (i.e., the same line) is: 4x + 6y – 16 = 0

This equation is simply a multiple of the original one, which means both equations represent the same line geometrically.

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Question 7:

Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:

Given the equations: x – y + 1 = 0 and 3x + 2y – 12 = 0

To draw the graphs of these equations and determine the triangle formed by them and the x-axis:

1. x – y + 1 = 0 can be rewritten as y = x + 1

2. 3x + 2y – 12 = 0 can be rewritten as y = (12 – 3x)/2

These lines intersect and also meet the x-axis at certain points, forming a triangle.

Vertices of the triangle are: (–1, 0), (4, 0), and (2, 3)

These are the points where the lines intersect each other and the x-axis.

You can now plot these points, draw the lines, and shade the triangular region enclosed by them.

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EXERCISE 3.2

1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14 x – y = 4

Solution

Given Equations:

1. x + y = 14

2. x – y = 4

Step 1: Solve one equation for one variable

From equation (2):

x – y = 4

Add y to both sides: x = y + 4

Step 2: Substitute this value of x into equation (1)

x + y = 14

Replace x with (y + 4):

(y + 4) + y = 14

2y + 4 = 14

Subtract 4 from both sides:

2y = 10

Divide both sides by 2:

y = 5

Step 3: Substitute the value of y back into the expression for x

x = y + 4

x = 5 + 4

x = 9

Final Answer: x = 9, y = 5

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(ii) s – t = 3, (s / 3) + (t / 2) = 6

Solution

Given Equations:

1. s – t = 3

2. (s / 3) + (t / 2) = 6

Step 1: Solve one equation for one variable

From equation (2): s – t = 3

Add “t” to both sides: s = t + 3

Step 2: Substitute this value of s into equation (1)

Replace s with (t + 3) in equation (1):

(t + 3)/3 + t/2 = 6

To simplify, find a common denominator for the fractions (LCM of 3 and 2 is 6):

[(t + 3) × 2]/6 + [t × 3]/6 = 6

(2t + 6 + 3t)/6 = 6

(5t + 6)/6 = 6

Multiply both sides by 6:

5t + 6 = 36

Subtract 6 from both sides:

5t = 30

Divide by 5:

t = 6

Step 3: Substitute the value of t back into s = t + 3

s = 6 + 3

s = 9

Final Answer: s = 9, t = 6

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(iii) 3x – y = 3, 9x – 3y = 9

Solution:

Given Equations:

1. 3x – y = 3

2. 9x – 3y = 9

Step 1: Solve one equation for one variable

Take equation (1):
3x – y = 3
Rearrange it to solve for y:
y = 3x – 3

Step 2: Substitute this value of y into equation (2)

Substitute y = 3x – 3 into equation (2):
9x – 3y = 9
9x – 3(3x – 3) = 9
9x – (9x – 9) = 9
9x – 9x + 9 = 9
9 = 9 (This is always true)

Conclusion:

Since the equation simplifies to a true statement (9 = 9), this means the two equations represent the same line, and there are infinitely many solutions.

The solution can be written as: y = 3x – 3, where x can be any real number.

Answer:
y = 3x – 3, where x can take any real value (infinitely many solutions).

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(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

Solution:

Given Equations:

1. 0.2x + 0.3y = 1.3

2. 0.4x + 0.5y = 2.3

Step 1: Solve one equation for one variable

Take equation (1):
0.2x + 0.3y = 1.3
Subtract 0.3y from both sides:
0.2x = 1.3 – 0.3y
Divide both sides by 0.2:
x = (1.3 – 0.3y) / 0.2

Simplify:
x = 6.5 – 1.5y

Step 2: Substitute this value of x into equation (2)

Substitute x = 6.5 – 1.5y into equation (2):

0.4x + 0.5y = 2.3
0.4(6.5 – 1.5y) + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
2.6 – 0.1y = 2.3

Subtract 2.6 from both sides:
–0.1y = –0.3
Divide both sides by –0.1:
y = 3

Step 3: Substitute the value of y back into the expression for x

x = 6.5 – 1.5y
x = 6.5 – 1.5(3)
x = 6.5 – 4.5
x = 2

Answer: x = 2, y = 3

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(v) √2x + √3y = 0, √3x – √8y = 0

Solution:

Given Equations:

1. √2x + √3y = 0

2. √3x – √8y = 0

Step 1: Solve one equation for one variable

From equation (1):
√2x + √3y = 0
Solve for x:
√2x = –√3y
x = (–√3y) / √2

Step 2: Substitute into equation (2)

Substitute x = (–√3y) / √2 into equation (2):

√3x – √8y = 0
√3((–√3y)/√2) – √8y = 0

Now simplify:
–(√3·√3y)/√2 – √8y = 0
–(3y)/√2 – √8y = 0

Recall that √8 = 2√2:
So we have:
–(3y)/√2 – 2√2y = 0

Now write both terms with a common denominator (√2):
–(3y)/√2 – (2√2·y) = –(3y)/√2 – (4y/√2)
= –(3y + 4y)/√2
= –7y/√2 = 0

Multiply both sides by √2:
–7y = 0
y = 0

Step 3: Substitute y = 0 back into x = (–√3y)/√2

x = (–√3·0)/√2 = 0

Answer: x = 0, y = 0

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(vi) (3x / 2) – (5y / 3) = –2, (x / 3) + (y / 2) = 13 / 6

Solution

Given Equations:

1. (3x / 2) – (5y / 3) = –2

2. (x / 3) + (y / 2) = 13 / 6

Step 1: Solve one equation for one variable

Take equation (2):
(x / 3) + (y / 2) = 13 / 6

Find the least common denominator (LCD is 6), and multiply the entire equation by 6 to eliminate fractions:

6 × (x / 3) + 6 × (y / 2) = 6 × (13 / 6)
2x + 3y = 13

Now solve for x:
2x = 13 – 3y
x = (13 – 3y) / 2

Step 2: Substitute into equation (1)

Original equation (1):
(3x / 2) – (5y / 3) = –2

Substitute x = (13 – 3y) / 2:

(3/2) × [(13 – 3y)/2] – (5y / 3) = –2
= (3(13 – 3y)) / 4 – (5y / 3) = –2
= (39 – 9y) / 4 – (5y / 3) = –2

Now find the LCM of 4 and 3 (which is 12), and eliminate fractions:

Multiply the entire equation by 12:

12 × [(39 – 9y)/4] – 12 × (5y / 3) = 12 × (–2)
3 × (39 – 9y) – 4 × 5y = –24
117 – 27y – 20y = –24
117 – 47y = –24

Subtract 117 from both sides:

–47y = –141
Divide by –47:

y = 3

Step 3: Substitute y = 3 into x = (13 – 3y) / 2

x = (13 – 3×3) / 2
x = (13 – 9) / 2
x = 4 / 2
x = 2

Answer: x = 2, y = 3

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Question 2

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution

1. 2x + 3y = 11

2. 2x - 4y = -24

Step 1: Eliminate one variable
Subtract equation (2) from equation (1):

(2x + 3y) - (2x - 4y) = 11 - (-24)
2x + 3y - 2x + 4y = 11 + 24
7y = 35
y = 5

Step 2: Substitute y = 5 into one equation to find x
Using equation (1):
2x + 3(5) = 11
2x + 15 = 11
2x = -4
x = -2

So the solution is:
x = -2, y = 5

Step 3: Find the value of m in y = mx + 3
Substitute x = -2 and y = 5 into the equation:

5 = m(-2) + 3
5 = -2m + 3
2 = -2m
m = -1

Answer: x = -2, y = 5, m = -1

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Question 3

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution:

Given:
The difference between two numbers is 26, and one number is three times the other.

Let the numbers be x and y, where x > y.

From the given information:

1. x - y = 26

2. x = 3y

Step 1: Substitute Equation 2 into Equation 1

Substitute x = 3y into the first equation:

3y - y = 26
2y = 26
y = 13

Step 2: Find x using y = 13

x = 3y = 3 × 13 = 39

Answer: x = 39, y = 13

So, the two numbers are 39 and 13.

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Q3

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:
Let the two angles be x and y (in degrees), where x > y.

From the given information:

1. x - y = 18

2. x + y = 180 (since the angles are supplementary)

Step 1: Use substitution
From equation 1:
x = y + 18

Now substitute this into equation 2:

(y + 18) + y = 180
2y + 18 = 180
2y = 162
y = 81

Step 2: Find x using y = 81

x = y + 18 = 81 + 18 = 99

Answer: x = 99 degrees, y = 81 degrees

So, the two angles are 99° and 81°.

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Q3

(iii) The coach of a cricket team buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and each ball.

Solution:

Let the cost of one bat be x (in ₹)
Let the cost of one ball be y (in ₹)

From the given information:

1. 7x + 6y = 3800

2. 3x + 5y = 1750

Step 1: Solve one equation for x
From equation 2:
3x + 5y = 1750
⇒ 3x = 1750 - 5y
⇒ x = (1750 - 5y) / 3

Step 2: Substitute into equation 1

Substitute x in equation 1:

7[(1750 - 5y) / 3] + 6y = 3800
(12250 - 35y) / 3 + 6y = 3800

Multiply through by 3 to eliminate the denominator:

12250 - 35y + 18y = 11400
12250 - 17y = 11400
-17y = 11400 - 12250
-17y = -850
y = 50

Step 3: Find x using y = 50

x = (1750 - 5×50) / 3
x = (1750 - 250) / 3
x = 1500 / 3
x = 500

Final Answer:
x = ₹500 (cost of one bat)
y = ₹50 (cost of one ball)

So, the cost of each bat is ₹500 and each ball is ₹50.

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Q3

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ` 105 and for a journey of 15 km, the charge paid is ` 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Solution:

Let the fixed charge be x (in ₹)
Let the charge per km be y (in ₹)

From the given information:

1. x + 10y = 105

2. x + 15y = 155

Step 1: Subtract the equations

Subtract equation 1 from equation 2:

(x + 15y) - (x + 10y) = 155 - 105
x + 15y - x - 10y = 50
5y = 50
y = 10

Step 2: Find x using y = 10

Use equation 1:
x + 10y = 105
x + 10×10 = 105
x + 100 = 105
x = 5

Step 3: Cost for 25 km

Total charge = x + 25y
= 5 + 25×10
= 5 + 250
= ₹255

Final Answer:
Fixed charge = ₹5
Charge per km = ₹10
Cost for 25 km = ₹255

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Q3

(v) A fraction becomes , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction

Solution:

Let the numerator be x
Let the denominator be y

From the given conditions:

1. (x + 2) / (y + 2) = 9 / 11

2. (x + 3) / (y + 3) = 5 / 6

Step 1: Cross-multiply both equations

From equation 1:
11(x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x - 9y = -4 → (Equation A)

From equation 2:
6(x + 3) = 5(y + 3)
6x + 18 = 5y + 15
6x - 5y = -3 → (Equation B)

Step 2: Solve by substitution

From Equation B:
6x - 5y = -3
⇒ 6x = 5y - 3
⇒ x = (5y - 3) / 6

Substitute into Equation A:

11[(5y - 3)/6] - 9y = -4
(55y - 33)/6 - 9y = -4
Now eliminate the denominator by multiplying the whole equation by 6:

55y - 33 - 54y = -24
(55y - 54y) - 33 = -24
y - 33 = -24
y = 9

Step 3: Find x using y = 9

Use Equation B:
6x - 5y = -3
6x - 5×9 = -3
6x - 45 = -3
6x = 42
x = 7

Final Answer:
Numerator = 7
Denominator = 9

So, the fraction is 7/9

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Q3

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution

Let the present age of Jacob be x (in years)
Let the present age of his son be y (in years)

From the given information:

1. Five years hence:
x + 5 = 3(y + 5)

2. Five years ago:
x - 5 = 7(y - 5)

Step 1: Simplify both equations

From equation 1:
x + 5 = 3(y + 5)
x + 5 = 3y + 15
x - 3y = 10 → Equation A

From equation 2:
x - 5 = 7(y - 5)
x - 5 = 7y - 35
x - 7y = -30 → Equation B

Step 2: Solve using substitution

From Equation A:
x = 3y + 10

Substitute into Equation B:

(3y + 10) - 7y = -30
3y + 10 - 7y = -30
-4y + 10 = -30
-4y = -40
y = 10

Step 3: Find x using y = 10

x = 3y + 10 = 3×10 + 10 = 40

Final Answer:
Jacob’s age = 40 years
His son’s age = 10 years

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Exercise 3.3

Question 1

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x – 3y = 4

Solution

We are given the system of equations:

1. x + y = 5

2. 2x – 3y = 4

Let's solve this first using the elimination method:

Step 1: Multiply the first equation by 2 so we can eliminate x later:
2(x + y) = 2(5)
→ 2x + 2y = 10

Step 2: Now subtract the second equation from this result:
(2x + 2y) – (2x – 3y) = 10 – 4
2x + 2y – 2x + 3y = 6
5y = 6
→ y = 6/5

Step 3: Substitute y = 6/5 into the first equation:
x + 6/5 = 5
→ x = 5 – 6/5 = 25/5 – 6/5 = 19/5

Answer: x = 19/5 and y = 6/5

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Q1

(ii) 3x + 4y = 10 and 2x – 2y = 2

Solution:

Let's solve the system of equations:

1. 3x + 4y = 10

2. 2x – 2y = 2

First: Elimination Method

Step 1: We'll eliminate y. Multiply the second equation by 2 to match y terms with the first:
2(2x – 2y) = 2(2) → 4x – 4y = 4

Now add this to the first equation:
(3x + 4y) + (4x – 4y) = 10 + 4
→ 7x = 14
→ x = 2

Step 2: Plug x = 2 into the first equation:
3(2) + 4y = 10
6 + 4y = 10
→ 4y = 4
→ y = 1

Now: Substitution Method

Step 1: From the second equation:
2x – 2y = 2
→ x – y = 1
→ x = y + 1

Step 2: Substitute into the first equation:
3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y + 3 = 10
→ 7y = 7
→ y = 1

Step 3: Plug y = 1 into x = y + 1 → x = 2

Final answer: x = 2, y = 1

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Q1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Solution:

We are given the system:

1. 3x – 5y – 4 = 0

2. 9x = 2y + 7

First: Elimination Method

Step 1: Rearrange both equations

Equation 1:
3x – 5y = 4 (move 4 to the other side)

Equation 2:
9x – 2y = 7 (move 2y to the left)

Now eliminate x.
Multiply equation 1 by 3 so we can match 9x in both equations:

3(3x – 5y) = 3(4) → 9x – 15y = 12
Equation 2 is: 9x – 2y = 7

Now subtract:
(9x – 15y) – (9x – 2y) = 12 – 7
9x – 15y – 9x + 2y = 5
→ –13y = 5
→ y = –5/13

Step 2: Substitute y = –5/13 into equation 1:
3x – 5(–5/13) = 4
3x + 25/13 = 4
→ 3x = 4 – 25/13 = (52 – 25)/13 = 27/13
→ x = 9/13

So the solution is x = 9/13 and y = –5/13

Now: Substitution Method

Step 1: From equation 2:
9x = 2y + 7
→ x = (2y + 7)/9

Step 2: Substitute into equation 1:
3((2y + 7)/9) – 5y = 4
(6y + 21)/9 – 5y = 4
→ (6y + 21 – 45y)/9 = 4
→ (–39y + 21)/9 = 4
Multiply both sides by 9:
–39y + 21 = 36
→ –39y = 15
→ y = –5/13

Then x = (2(–5/13) + 7)/9
= (–10/13 + 91/13)/9 = 81/13 ÷ 9 = 9/13

Answer: Same result: x = 9/13 and y = –5/13

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Q1

(iv) x/2 + 2y/3 = -1 and x - y/3 = 3

Solution:

We are given:

1. x/2 + 2y/3 = -1

2. x - y/3 = 3

First: Elimination Method

Step 1: Eliminate denominators by multiplying both equations:

Equation 1: Multiply through by 6
6(x/2 + 2y/3) = 6(–1)
→ 3x + 4y = –6

Equation 2: Multiply through by 3
3(x - y/3) = 3(3)
→ 3x – y = 9

Now we have:

1. 3x + 4y = –6

2. 3x – y = 9

Step 2: Subtract the second equation from the first:
(3x + 4y) – (3x – y) = –6 – 9
3x + 4y – 3x + y = –15
5y = –15
→ y = –3

Step 3: Substitute y = –3 into equation 2:
3x – (–3) = 9
3x + 3 = 9
→ 3x = 6
→ x = 2

So the solution is x = 2 and y = –3

Now: Substitution Method

Step 1: Take equation 2:
x – y/3 = 3
→ x = 3 + y/3

Step 2: Substitute into equation 1:
(3 + y/3)/2 + 2y/3 = –1

Find a common denominator and simplify:

First term: (3 + y/3)/2 = (9 + y)/6
Second term: 2y/3

So: (9 + y)/6 + 2y/3 = –1
Convert 2y/3 to same denominator: 4y/6

Now: (9 + y + 4y)/6 = –1
→ (9 + 5y)/6 = –1
→ 9 + 5y = –6
→ 5y = –15
→ y = –3

Now plug back into x = 3 + y/3
x = 3 – 1 = 2

Answer: x = 2, y = –3 confirmed.

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Question 2:

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Solution:

We are given a word problem and asked to:

1. Form equations using variables

2. Solve using elimination method

Step 1: Let the fraction be x/y

Let x = numerator, y = denominator

First condition:

If we add 1 to the numerator and subtract 1 from the denominator, the fraction becomes 1.

That means:
(x + 1)/(y – 1) = 1
→ x + 1 = y – 1
→ x – y + 2 = 0  → (Equation 1)

Second condition:

If we add 1 to the denominator only, the fraction becomes 1/2.

So:
x / (y + 1) = 1/2
→ 2x = y + 1
→ 2x – y – 1 = 0  → (Equation 2)

Step 2: Solve by elimination method

We now solve:

1. x – y + 2 = 0

2. 2x – y – 1 = 0

Subtract equation 1 from equation 2:
(2x – y – 1) – (x – y + 2) = 0
2x – y – 1 – x + y – 2 = 0
→ x – 3 = 0
→ x = 3

Now plug into equation 1:
3 – y + 2 = 0
→ 5 – y = 0
→ y = 5

Final Answer: x = 3, y = 5 → The fraction is 3/5
Equations:
x – y + 2 = 0
2x – y – 1 = 0
Solved by elimination method.

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Q2

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution:

We are given:

• Five years ago, Nuri was three times as old as Sonu.

• Ten years later, Nuri will be twice as old as Sonu.

• Let x = Nuri’s current age, y = Sonu’s current age.

Step 1: Form equations

Condition 1: Five years ago
Nuri’s age = x – 5
Sonu’s age = y – 5
Given: x – 5 = 3(y – 5)

Simplify:
x – 5 = 3y – 15
→ x – 3y + 10 = 0  → (Equation 1)

Condition 2: Ten years later
Nuri’s age = x + 10
Sonu’s age = y + 10
Given: x + 10 = 2(y + 10)

Simplify:
x + 10 = 2y + 20
→ x – 2y – 10 = 0  → (Equation 2)

Step 2: Solve using elimination method

We solve:

1. x – 3y + 10 = 0

2. x – 2y – 10 = 0

Subtract equation 2 from equation 1:

(x – 3y + 10) – (x – 2y – 10) = 0
x – 3y + 10 – x + 2y + 10 = 0
–y + 20 = 0
→ y = 20

Substitute y = 20 into equation 1:

x – 3(20) + 10 = 0
x – 60 + 10 = 0
→ x = 50

Final Answer:
Nuri’s age = 50 years
Sonu’s age = 20 years
Equations:
x – 3y + 10 = 0
x – 2y – 10 = 0
Solved by elimination method.

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Q2

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:

Let the two-digit number be 10x + y, where:

• x = tens digit

• y = units digit

Step 1: Form equations

Condition 1: The sum of the digits is 9
→ x + y = 9  → (Equation 1)

Condition 2: Nine times the number is twice the number formed by reversing the digits
Original number = 10x + y
Reversed number = 10y + x

Given:
9(10x + y) = 2(10y + x)

Simplify both sides:
90x + 9y = 20y + 2x
→ 90x – 2x + 9y – 20y = 0
→ 88x – 11y = 0
→ 8x – y = 0  → (Equation 2)

Step 2: Solve using elimination method

We now solve:

1. x + y = 9

2. 8x – y = 0

Add both equations:
(x + y) + (8x – y) = 9
x + y + 8x – y = 9
→ 9x = 9
→ x = 1

Now substitute x = 1 into equation 1:
1 + y = 9 → y = 8

Final step:

The number is 10x + y = 10(1) + 8 = 18

Final Answer:
Equations: x + y = 9, 8x – y = 0
x = 1, y = 8 → Number = 18
Solved by elimination method.

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Q2

(iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of ` 50 and ` 100 she received.

Solution:

Step 1: Let the variables represent what we want to find

Let
x = number of ₹50 notes
y = number of ₹100 notes

Step 2: Form the equations

From the question:

1. Meena got 25 notes in total
So, x + y = 25 … (1)

2. The total amount of money is ₹2000
So, 50x + 100y = 2000

Divide every term by 50: x + 2y = 40 … (2)

Step 3: Solve the system of equations

We now solve the two equations:
x + y = 25 … (1)
x + 2y = 40 … (2)

Subtract equation (1) from equation (2):
(x + 2y) - (x + y) = 40 - 25
x + 2y - x - y = 15
y = 15

Now substitute y = 15 into equation (1):
x + 15 = 25
x = 10

Final Answer:
Number of ₹50 notes = 10
Number of ₹100 notes = 15

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Q2

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy paid ` 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

Step 1: Let the variables represent what we want to find

Let
x = fixed charge for the first 3 days (in ₹)
y = additional charge per extra day (in ₹)

Step 2: Form the equations

From the question:

• Saritha kept the book for 7 days.
  So she paid: fixed charge (x) + 4 extra days (4y)
  Therefore,
  x + 4y = ₹27 … (1)

• Susy kept the book for 5 days.
  So she paid: fixed charge (x) + 2 extra days (2y)
  Therefore,
  x + 2y = ₹21 … (2)

Step 3: Solve the system of equations

We now solve the two equations:
x + 4y = 27 … (1)
x + 2y = 21 … (2)

Subtract equation (2) from equation (1):
(x + 4y) - (x + 2y) = 27 - 21
x + 4y - x - 2y = 6
2y = 6
y = 3

Now substitute y = 3 into equation (2):
x + 2(3) = 21
x + 6 = 21
x = 15

Final Answer:
Fixed charge = ₹15
Additional charge per extra day = ₹3