MAK LABS — Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

1Form 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.

Let the number of girls be x and the number of boys be y.

x + y = 10 ... (1)

x – y = 4 ... (2)

For x + y = 10:

x	0	5	10
y	10	5	0

For x – y = 4:

x	4	5	7
y	0	1	3

The lines intersect at (7, 3).

The two lines intersect at the point (7, 3). So, x = 7 and y = 3.

Number of girls = 7 and number of boys = 3

(ii) 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.

Let the cost of one pencil be ₹ x and the cost of one pen be ₹ y.

5x + 7y = 50 ... (1)

7x + 5y = 46 ... (2)

For 5x + 7y = 50:

x	3	10	–4
y	5	0	10

For 7x + 5y = 46:

x	3	8	–2
y	5	–2	12

The lines intersect at (3, 5).

The two lines intersect at the point (3, 5). So, x = 3 and y = 5.

Cost of one pencil = ₹ 3 and cost of one pen = ₹ 5

2On comparing the ratios a₁a₂, b₁b₂ and c₁c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0 ; 7x + 6y – 9 = 0 (ii) 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0 ; 2x – y + 9 = 0

(i) 5x – 4y + 8 = 0 ; 7x + 6y – 9 = 0

a₁a₂ = 57 ; b₁b₂ = –46 = –23

Here, a₁a₂ ≠ b₁b₂.

The lines intersect at a point.

(ii) 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0

a₁a₂ = 918 = 12 ; b₁b₂ = 36 = 12 ; c₁c₂ = 1224 = 12

Here, a₁a₂ = b₁b₂ = c₁c₂.

The lines are coincident.

(iii) 6x – 3y + 10 = 0 ; 2x – y + 9 = 0

a₁a₂ = 62 = 3 ; b₁b₂ = –3–1 = 3 ; c₁c₂ = 109

Here, a₁a₂ = b₁b₂ ≠ c₁c₂.

The lines are parallel.

3On comparing the ratios a₁a₂, b₁b₂ and c₁c₂, find out whether the following pair of linear equations are consistent, or inconsistent. (i) 3x + 2y = 5 ; 2x – 3y = 7 (ii) 2x – 3y = 8 ; 4x – 6y = 9 (iii) 32x + 53y = 7 ; 9x – 10y = 14 (iv) 5x – 3y = 11 ; –10x + 6y = –22 (v) 43x + 2y = 8 ; 2x + 3y = 12

(i) 3x + 2y = 5 ; 2x – 3y = 7

a₁a₂ = 32 ; b₁b₂ = 2–3 ⇒ a₁a₂ ≠ b₁b₂

The lines intersect at one point (unique solution).

The pair is consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9

a₁a₂ = 24 = 12 ; b₁b₂ = –3–6 = 12 ; c₁c₂ = 89 ⇒ a₁a₂ = b₁b₂ ≠ c₁c₂

The lines are parallel (no solution).

The pair is inconsistent.

(iii) 32x + 53y = 7 ; 9x – 10y = 14

a₁a₂ = 3/29 = 16 ; b₁b₂ = 5/3–10 = –16 ⇒ a₁a₂ ≠ b₁b₂

The lines intersect at one point (unique solution).

The pair is consistent.

(iv) 5x – 3y = 11 ; –10x + 6y = –22

a₁a₂ = 5–10 = –12 ; b₁b₂ = –36 = –12 ; c₁c₂ = 11–22 = –12 ⇒ a₁a₂ = b₁b₂ = c₁c₂

The lines are coincident (infinitely many solutions).

The pair is consistent (dependent).

(v) 43x + 2y = 8 ; 2x + 3y = 12

a₁a₂ = 4/32 = 23 ; b₁b₂ = 23 ; c₁c₂ = 812 = 23 ⇒ a₁a₂ = b₁b₂ = c₁c₂

The lines are coincident (infinitely many solutions).

The pair is consistent (dependent).

4Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x – 3y = 16 (iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 (iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

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

a₁a₂ = 12 ; b₁b₂ = 12 ; c₁c₂ = 510 = 12

All three ratios are equal. So, the lines are coincident and the pair is consistent.

For x + y = 5:

x	0	2	5
y	5	3	0

Both equations give the same line (coincident lines).

Every point on the line is a solution, e.g., (0, 5), (2, 3), (5, 0).

Consistent; infinitely many solutions — any x, with y = 5 – x

(ii) x – y = 8, 3x – 3y = 16

a₁a₂ = 13 ; b₁b₂ = –1–3 = 13 ; c₁c₂ = 816 = 12

Here, a₁a₂ = b₁b₂ ≠ c₁c₂. So, the lines are parallel.

The pair is inconsistent (no solution).

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

a₁a₂ = 24 = 12 ; b₁b₂ = 1–2 ⇒ a₁a₂ ≠ b₁b₂

So, the lines intersect and the pair is consistent.

For 2x + y = 6:

x	0	2	3
y	6	2	0

For 4x – 2y = 4:

x	0	1	2
y	–2	0	2

The lines intersect at (2, 2).

Consistent; solution x = 2, y = 2

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

a₁a₂ = 24 = 12 ; b₁b₂ = –2–4 = 12 ; c₁c₂ = –2–5 = 25

Here, a₁a₂ = b₁b₂ ≠ c₁c₂. So, the lines are parallel.

The pair is inconsistent (no solution).

5Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Let the length be x m and the width be y m.

x = y + 4, i.e., x – y = 4 ... (1)

Half perimeter = x + y = 36 ... (2)

Adding (1) and (2): 2x = 40, so x = 20.

From (2): y = 36 – 20 = 16.

The graphs of these two lines intersect at the point (20, 16).

Length = 20 m and width = 16 m

6Given 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 (ii) parallel lines (iii) coincident lines

(i) Intersecting lines

We need a₁a₂ ≠ b₁b₂.

One such equation is 3x + 2y – 7 = 0

(ii) Parallel lines

We need a₁a₂ = b₁b₂ ≠ c₁c₂.

One such equation is 4x + 6y – 9 = 0

(iii) Coincident lines

We need a₁a₂ = b₁b₂ = c₁c₂.

One such equation is 6x + 9y – 24 = 0

Many other answers are possible. Check your answer using the ratio conditions above.

7Draw 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.

For x – y + 1 = 0:

x	–1	0	2
y	0	1	3

For 3x + 2y – 12 = 0:

x	0	2	4
y	6	3	0

The shaded region is the triangle ABC formed by the two lines and the x-axis.

The line x – y + 1 = 0 meets the x-axis at A(–1, 0).

The line 3x + 2y – 12 = 0 meets the x-axis at B(4, 0).

The two lines intersect each other at C(2, 3).

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

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

(i) x + y = 14 ; x – y = 4

From the second equation: x = y + 4.

(y + 4) + y = 14 ⇒ 2y = 10 ⇒ y = 5

x = 5 + 4 = 9

Solution x = 9, y = 5

(ii) s – t = 3 ; s3 + t2 = 6

From the first equation: s = t + 3.

t + 33 + t2 = 6 ⇒ 2(t + 3) + 3t = 36 ⇒ 5t = 30 ⇒ t = 6

s = 6 + 3 = 9

Solution s = 9, t = 6

(iii) 3x – y = 3 ; 9x – 3y = 9

From the first equation: y = 3x – 3.

9x – 3(3x – 3) = 9 ⇒ 9x – 9x + 9 = 9 ⇒ 9 = 9

This statement is true for all values of x. So, the equations have infinitely many solutions.

Infinitely many solutions y = 3x – 3, where x is any real number

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

Multiplying both equations by 10: 2x + 3y = 13 and 4x + 5y = 23.

From the first: x = 13 – 3y2.

4 × 13 – 3y2 + 5y = 23 ⇒ 26 – 6y + 5y = 23 ⇒ y = 3

x = 13 – 92 = 2

Solution x = 2, y = 3

(v) √2x + √3y = 0 ; √3x – √8y = 0

From the first equation: x = –√3y√2.

√3 × –√3y√2 – √8y = 0 ⇒ –3y√2 – 2√2y = 0 ⇒ –3y – 4y = 0 ⇒ y = 0

x = –√3 × 0√2 = 0

Solution x = 0, y = 0

(vi) 3x2 – 5y3 = –2 ; x3 + y2 = 136

Multiplying the first by 6: 9x – 10y = –12. Multiplying the second by 6: 2x + 3y = 13.

From 2x + 3y = 13: x = 13 – 3y2.

9 × 13 – 3y2 – 10y = –12 ⇒ 117 – 27y – 20y = –24 ⇒ 47y = 141 ⇒ y = 3

x = 13 – 92 = 2

Solution x = 2, y = 3

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

From 2x + 3y = 11: 2x = 11 – 3y.

(11 – 3y) – 4y = –24 ⇒ –7y = –35 ⇒ y = 5

2x = 11 – 15 = –4 ⇒ x = –2

Now, put x = –2 and y = 5 in y = mx + 3:

5 = –2m + 3 ⇒ 2m = –2 ⇒ m = –1

x = –2, y = 5 and m = –1

3Form 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.

Let the larger number be x and the smaller be y.

x – y = 26 ; x = 3y

3y – y = 26 ⇒ y = 13 ; x = 3 × 13 = 39

The numbers are 39 and 13

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

Let the larger angle be x° and the smaller be y°.

x + y = 180 ; x – y = 18 ⇒ x = y + 18

(y + 18) + y = 180 ⇒ 2y = 162 ⇒ y = 81 ; x = 99

The angles are 99° and 81°

(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.

Let the cost of a bat be ₹ x and of a ball be ₹ y.

7x + 6y = 3800 ; 3x + 5y = 1750

From the second: x = 1750 – 5y3.

7 × 1750 – 5y3 + 6y = 3800 ⇒ 12250 – 35y + 18y = 11400 ⇒ 17y = 850 ⇒ y = 50

x = 1750 – 2503 = 500

Cost of a bat = ₹ 500 and cost of a ball = ₹ 50

(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?

Let the fixed charge be ₹ x and the charge per km be ₹ y.

x + 10y = 105 ; x + 15y = 155

From the first: x = 105 – 10y.

105 – 10y + 15y = 155 ⇒ 5y = 50 ⇒ y = 10 ; x = 105 – 100 = 5

Charge for 25 km = x + 25y = 5 + 250 = 255

Fixed charge = ₹ 5, charge per km = ₹ 10, fare for 25 km = ₹ 255

(v) A fraction becomes 911, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 56. Find the fraction.

Let the fraction be xy.

x + 2y + 2 = 911 ⇒ 11x + 22 = 9y + 18 ⇒ 11x – 9y = –4

x + 3y + 3 = 56 ⇒ 6x + 18 = 5y + 15 ⇒ 6x – 5y = –3

From the second: x = 5y – 36.

11 × 5y – 36 – 9y = –4 ⇒ 55y – 33 – 54y = –24 ⇒ y = 9

x = 45 – 36 = 7

The fraction is 79

(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?

Let Jacob’s present age be x years and his son’s be y years.

x + 5 = 3(y + 5) ⇒ x – 3y = 10

x – 5 = 7(y – 5) ⇒ x – 7y = –30

From the first: x = 3y + 10.

3y + 10 – 7y = –30 ⇒ –4y = –40 ⇒ y = 10 ; x = 40

Jacob is 40 years old and his son is 10 years old

1Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

Elimination method: Multiply the first equation by 3 and add.

3x + 3y = 15 ; 2x – 3y = 4 ⇒ 5x = 19 ⇒ x = 195

y = 5 – 195 = 65

Substitution method: From the first, y = 5 – x.

2x – 3(5 – x) = 4 ⇒ 5x = 19 ⇒ x = 195 ; y = 5 – 195 = 65

Solution x = 195, y = 65

(ii) 3x + 4y = 10 and 2x – 2y = 2

Elimination method: Multiply the second equation by 2 and add.

3x + 4y = 10 ; 4x – 4y = 4 ⇒ 7x = 14 ⇒ x = 2

2(2) – 2y = 2 ⇒ y = 1

Substitution method: From the second, x = 1 + y.

3(1 + y) + 4y = 10 ⇒ 7y = 7 ⇒ y = 1 ; x = 2

Solution x = 2, y = 1

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

Elimination method: Write as 3x – 5y = 4 and 9x – 2y = 7. Multiply the first by 3 and subtract.

9x – 15y = 12 ; 9x – 2y = 7 ⇒ –13y = 5 ⇒ y = –513

3x = 4 + 5y = 4 – 2513 = 2713 ⇒ x = 913

Substitution method: From 3x – 5y = 4, x = 4 + 5y3.

9 × 4 + 5y3 – 2y = 7 ⇒ 12 + 15y – 2y = 7 ⇒ 13y = –5 ⇒ y = –513 ; x = 913

Solution x = 913, y = –513

(iv) x2 + 2y3 = –1 and x – y3 = 3

Clearing fractions: 3x + 4y = –6 and 3x – y = 9.

Elimination method: Subtract the second from the first.

5y = –15 ⇒ y = –3 ; 3x = 9 + y = 6 ⇒ x = 2

Substitution method: From 3x – y = 9, y = 3x – 9.

3x + 4(3x – 9) = –6 ⇒ 15x = 30 ⇒ x = 2 ; y = 6 – 9 = –3

Solution x = 2, y = –3

2Form 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 12 if we only add 1 to the denominator. What is the fraction?

Let the fraction be xy.

x + 1y – 1 = 1 ⇒ x – y = –2 ... (1)

xy + 1 = 12 ⇒ 2x – y = 1 ... (2)

Subtracting (1) from (2): x = 3. Then y = x + 2 = 5.

The fraction is 35

(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?

Let Nuri’s present age be x years and Sonu’s be y years.

x – 5 = 3(y – 5) ⇒ x – 3y = –10 ... (1)

x + 10 = 2(y + 10) ⇒ x – 2y = 10 ... (2)

Subtracting (1) from (2): y = 20. Then x = 10 + 2y = 50.

Nuri is 50 years and Sonu is 20 years old

(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.

Let the tens digit be x and the units digit be y. The number = 10x + y.

x + y = 9 ... (1)

9(10x + y) = 2(10y + x) ⇒ 88x – 11y = 0 ⇒ 8x – y = 0 ... (2)

Adding (1) and (2): 9x = 9, so x = 1. Then y = 8.

The number is 18

(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.

Let the number of ₹ 50 notes be x and of ₹ 100 notes be y.

x + y = 25 ... (1)

50x + 100y = 2000 ⇒ x + 2y = 40 ... (2)

Subtracting (1) from (2): y = 15. Then x = 10.

₹ 50 notes = 10 and ₹ 100 notes = 15

(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.

Let the fixed charge be ₹ x and the charge for each extra day be ₹ y.

x + 4y = 27 ... (1) (7 days = 3 days + 4 extra days)

x + 2y = 21 ... (2) (5 days = 3 days + 2 extra days)

Subtracting (2) from (1): 2y = 6, so y = 3. Then x = 21 – 6 = 15.

Fixed charge = ₹ 15 and charge for each extra day = ₹ 3

3. Pair of Linear Equations in Two Variables

Exercise 3.1

Exercise 3.2

Exercise 3.3