Pioneer Education {The Best Way To Success} IIT JEE ... the following system of linear equations by...

Pioneer Education {The Best Way To Success} IIT – JEE /AIPMT/NTSE/Olympiads Classes

www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 5

Revision Question Bank

Pair of Linear Equations in Two Variables

1. Draw the graph of the equations x – y +1 = 0 and 3 x + 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 :

We have

x – y + 1 = 0

3x + 2y – 12 = 0

Graph of the equation x – y + 1 = 0

y = x + 1.

When x = 0, we have y = 1


Thus, we have the following table

x 0 –1

y 1 0

Plotting the points A(0, 1) and B(–1, 0) on the graph paper on a suitable scale and drawing a line joining

them.

Graph of the equation 3x + 2y – 12 = 0

2y = 12 – 3x



Thus, we have the following table

Plotting the points C(0, 6) and D(4, 0) on the same graph paper and joining them, we obtain the

following graph : –

x 0 4

y 6 0



It is evident from the graph that the two lines intersect at point P(2, 3). The area enclosed the lines and

x-axis is shown in the shaded region.

2. Solve the following system of linear equations

8v–3u = 5uv and 6v– 5u=–2uv.

Solution :

Clearly, the given equations are not linear in the variables u and v but can be reduced into linear

equations by an appropriate substitution.

If we put u = 0 in either of the two equations, we get v = 0.

Thus, u = 0 and v = 0 form one solution of the given system of equations.

To find the other solutions we assume that.

Since, u 0 and v 0 Therefore, uv 0 .

On dividing each of the given equations by uv, we get

3 3

5u v ..(i)

and 6 5

2u v ..(ii)

on putting 1

xu and

1

v = y, the above equations become

8x – 3v = 5 ...(iii)



and 6x – 5y = –2 ..(iv)

On multiplying Eq. (iii) by 3 and Eq. (iv) by 4, we get

24x – 9y = 15 ...(v)

and 24x – 20y = – 8 ...(vi)

On subtracting Eq. (vi) from Eq. (v), we get

11y = 23

23

y11

On putting y = 23

11 in Eq. (iii), we get

698x 5

11

60

8x 511

55 69

8x11

124 124 31

8x x11 8 11 22

Now, 31

x22

1 31 22u

u 22 31

and 23

y11

1 23 11v

v 11 23

Hence, the given system of equations has two solutions given by

(i) u = 0 and v = 0

(ii) 22 11

u and v31 23

3. Solve for x and y by using the method of elimination

0.4x + 0.3y =1.7and 0.7x– 0.2y = 0.8.

Solve for x and y by using the method of elimination

0.4x + 0.3y =1.7and 0.7x– 0.2y = 0.8.

Solution :

Given equations are 0.4x + 03y = 1.7 ... (i)

and 0.7x - 0.2y = 0.8 ... (ii)



On multiplying each one of the equations by 10, we get

4x + 3y= 17 ... (iii)

and 7x– 2y =8 ...(iv)

On multiplying Eq. (iii) by 2 and Eq. (iv) by 3, we get

8x + 6y = 34 ... (v)

and 21x – 6y = 24 ...(vi)

On adding Eqs. (v) and (vi), we get

29x = 58 x = 58

29 = 2

On putting x = 2 in Eq. (iv), we get

7 × 2 – 2y = 8 2y= 14 – 8

2y = 6 y = 3

x = 2 and y = 3

4. Solve the following system of linear equations by the method of cross-multiplication

2x– y – 3 = 0 and 4x + y– 3 = 0.

Solution :

The given system of equations is

2x – y – 3 = 0

and 4x + 3y –3 = 0

By cross-multiplication method, we have

x y

1 3 1 3 2 3 4 3

=

1

2 1 4 1

x y 1

3 3 6 12 2 4



x y 1

6 6 6

6 6

x 1 and y 16 6

Hence, the solution of the given system of equations is x = 1 and y = –1.

5. Solve the following system of equations in x and y

(a – b)x + (a + b)y =a2– 2ab– b2 and(a + b)(x + y) = a2 + b2.

Solve the following system of equations in x and y

(a – b)x + (a + b)y =a2– 2ab– b2 and(a + b)(x + y) = a2 + b2.

Solution :

The given system of equations may be written as

(a – b)x + (a + b)y – (a2 –2ab –b2) = 0

and (a + b)x + (a + b) y – (a2 + b) = 0

By cross-multiplication method, we have

2 2 2 2

x

a b a b a b a 2ab b

= 2 2 2 2

y

a b a b a b a 2ab b

=

2

1

a b a b a b

2 2 2 2

x

a b a b a b a 2ab b

2 2 2 2

y

a b a b a b a 2ab b

=

2

1

a b a b a b

2 2 2 2

x

a b [ a b a 2ab b ]

2 2 2 2

y

a b a 2ab b a b a b

=

1

a b a b a b



2

x

a b 2ab b

= 3 2 2 3 3 2 2 3

y

a a b 3ab b a ab a b b

=

1

a b 2b

2 2

x y 1

4ab 2b a b2b a b

22b a b

x a b2b a b

and

24ab 2aby

2b a b a b

Hence, the solution of the given system of equations is

x = a + b and y =2ab

a b

.

6. A Lending library has a fixed charge for the first three days and an additional charge for each day,

thereafter and Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she

kept for five days. Find the fixed charge and the charge for each extra day.

Solution :

Let fixed charge for first 3 days = Rs x

and additional charge for each day = Rs y

According to the question,

x + 4y = 27 ...(i)

and x + 2y = 21 ... (ii)

On subtracting Eq. (ii) from Eq. (i), we get

x 4y 27

x 2y 21

2y 6

6

y 32

On substituting the value of y in Eq, (i), we get

x + 4 × 3 = 27 x +12 = 27



x = 15

Hence, charge for first 3 days = Rs 15 and extra charge per day = Rs 3.

7. For the following system of equations, determine the value of k for which the given system of equations

has a unique solution. 2x + 3y– 5 = 0 and kx– 6y– 8 = 0

Solution :


2x + 3y – 5 = 0

and kx – 6y – 8 = 0

It is of the form a1x +b1y + c1 = 0

and a2x + b2y + c2 = 0

where, a1 = 2,b1 =3,c1 = –5

and a2 = k, b2 =-6, c2 = – 8

For a unique solution, we must have

1 1

2 2

a b

a b

2 3k 4

k 6

i.e.,

Hence, the given system of equations will have a unique solution for all real values of k other than – 4.

8. Five years ago, Ramesh was thrice of Shyam's age. Ten years later, Ramesh will be twice of Shyam's age.

How old are Ramesh and Shyam?

Solution :

Let the present age of Ramesh be x years

2nd the present age of Shyam be y years.

Five years ago, Ramesh’s age = (x – 5)years

Shyam’s age = (y – 5) years

Using the given information, we have

x – 5 = 3(y – 5)

x – 5 = 3y – 15

x – 3y = – 10 …(i)

Also, ten years, hence, Ramesh’s age = (x + 10) years

Shyam’s age = (y + 10) years

Using the given information, we have

(x + 10) = 2(y + 10)

x + 10 = 2y + 20



x – 2y = 10 ..(ii)

Subtracting equation (ii) from (i), we get

–3y + 2y = – 10 – 10

– y = – 20

y = 20

Putting y = 20 in equation (i), we get : –

x – 3(20) = – 10

x – 60 = – 10

x = – 10 + 60

x = 50

Hence, present age of Ramesh is 50 years 2nd present age of shyam is 20 years.

9. Father's age is 3 times the sum of ages of his two children. After 5 yr, his age will be twice the sum of

ages of the two children. Find the age of father.

Solution :

Let father's age be x year and let sum of ages of his two children by y year.

Condition I

x = 3y ...(i)

After 5 yr, age of father = (x + 5)

After 5 yr, sum of ages of two children =(y +10)

Condition II

x + 5 = 2(y+10) ...(ii)

On putting the value of x from Eq. (i) in Eq. (ii), we get

3y + 5 = 2(y+10)

3y+5 = 2y + 20

3y – 2y = 20 – 5 y=15

On putting the value of y in Eq. (i), we get

x = 3(15) = 45

Hence, age of father is 45 yr.

10. Draw the graph of the pair of equations 2x + y=4 and 2x - y=4. Write the vertices of the triangle formed

by these lines and the y-axis. Find the area of this triangle.

Solution :

The given pair of linear equations 2x + y =4 and 2x – y = 4.

Table for line 2x + y = 4y =4 – 2x



x 0 2

y = 4 – 2x 4 0

Points A(0, 4) B(2, 0)

Now, we plot all these points on a graph paper and join them to get a line AB.

and table for line 2x – y = 4 y = 2x – 4

x 0 2

y = 2x – 4 –4 0

Points C(0, 4) B(2, 0)

Now, we plot all these points on a graph paper and join them to get a line BC.

Graphical representation of both lines are given below

The vertices formed by a triangle are A (0, 4), B (2, 0) and C(0,–4).

Here, AC = OA + OC = 4 + 4 = 8

and OB =2

Area of ABC1

2× base ×height

= 1

2× AC × OB

= 1

2× 8 × 2 = 8 sq units.



Chapter Test {Polynomial} M:M: 40 M: Time: 40

Each Question carry 4 marks

1. A and B each have certain number of oranges. A says to B, if you give me 10 of your oranges, I will have

twice the .number of oranges left with you. B replies, if you give me 10 of your oranges, I will have the

same number of oranges as left with you. Find the number of oranges with A and B separately. [4]

Solution :

Let A has x oranges and B has y oranges. According to the question,

x + 10 = 2(y – 10)

x + 10 = 2y – 20

x – 2y = – 20 – 10

x – 2y = – 30 ..(i)

x – y = 20

On subtracting Eq. (ii) from Eq. (i), we get

x 2y 30

x y 20

y 50

y = 50

On putting y = 50 in Eq. (ii), we get

x – 50 = 20

x = 20 + 50

x = 70

Hence, A has 70 oranges and B has 50 oranges.

2. Draw the graph of the equations 4x – y = 4 and 4x + y = 12. Determine the vertices of the triangle formed

by the line representing these equations and the X-axis. Shade the triangular region, so formed. [4]

Solution :

The given equations are

4x – x = 4 ...(i)

and 4x + y –12 ...(ii)

From Eq. (i), we have

4x – y = 4 y = 4x – 4



Put x = 0, then y = 4 × 0 – 4 = – 4

Put x = 1, then y = 4 – 4 = 0

Put x = 2, then y =4 × 2 – 4 = 4

x 0 1 2

y –4 0 4

From Eq. (ii), we have

4x + y = 12 y = 12 – 4x

Put x = 3, then y = 12 – 4 × 3 = 0

Put x = 3, then y = 12 – 4 × 2 = 4

Put x = 4, then y = 12 – 4 × 4 = – 4

x 3 2 4

y 0 4 –4

Now, plotting the points (0, –4), (1, 0), (2, 4), (3, 0) and (4, –4) on the graph paper with a suitable scale and

drawing lines joining them equation wise, we obtain the graph of the lines represented by the equations 4x

– y = 4 and 4x + y = 12,

From the graph, we observe that points A(2, 4), B (1, 0) and C(3, 0) are the three vertices of the triangle

formed by the given lines and the X-axis.

3. An express train takes 1 h less than a passenger train to travel 132 km between Mysore and Bangalore

(without taking into consideration the time for which they stop at intermediate stations). If average

speed of the express train is 11 km/h more than average speed of the passenger train, then find the

speed of the express train. [4]

Solution :

Let average speed of the passenger train = xkm/h and average speed of the express train



= (x+11)km/h

Time taken to cover 132 km by passenger train

132

x h

and time taken to cover 132 km by express train

= 132

x 11h

According to the question,132 132

1x x 11

132 x 1 132x1

x x 11

132x + 1452 – 132x = x2 +11X

x2+11x – 1452 = 0

x2 +44x – 33x – 1452 = 0

[by splitting middle term]

(x – 33)(x +44) = 0

x – 33 = 0 and x + 44= 0

x = 33 and x = – 44 [impossible]

Hence, the speed of express tram is 33+11 i.e., 44 km/h.

4. A train travels 288 km at a uniform speed. If the speed had been 4 km/h more, if would have taken 1 h

less for the same journey. Find the speed of the train. [4]

Solution :

Let the speed of the train be x km/h.

Time taken to cover 288 km at x km/h =288

xh

Time taken to cover 288 km at (x + 4) km/h

=

288

x 4


288 288

x x 4

= 1

288 (x + 4) – 288x = x (x + 4)

288x + 1152 – 288x = x2 +4x



x2+4x – 1152 = 0

x2 +36x – 32x – 1152 = 0

[by splitting middle term]

x (x + 36) – 32 (x + 36) = 0

(x + 36)(x – 32) = 0

x + 36=0 or x – 32 = 0

x = – 36 or x = 32

x = 32

[ x 36 , as speed cannot be negative]

Hence, the speed of the train is 32 km/h.

5. Solve for x and y [4]

15 225

x y x y

40 5513

x y x y

x y and x y

Solution :

On putting 1 1

u and vx y x y

, then given equations become

15u + 22v = 5 ..(i)

and 40u + 55y = 13 ..(ii)

On multiplying Eq. (i) by 5 and Eq. (ii) by 2 and then subtract, we get

75u 110v 25

80u 110v 26

5u 1

1

u5

On putting 1

u5

in Eq. (i), we get

115 22v 5

5

22 v = 5 – 3



2 1

v v22 11

1 1 1 1

andx y 5 x y 11

1 1

u and v5 11

x – y = 5 ..(iii)

x + y = 11 ..(iv)

On solving Equations. (iii) and (iv), we get

x = 8,y = 3

x = 8 and y = 3 is the required solution.

6. Find the value of k, in which system equations x + 3y = 2 and 2x + ky = 8 has no solution. [4]

Solution :

Given lines are x + 3y = 2 and 2x + ky = 8

On comparing with ax + by = c, we get

a1 =1, b1 = 3, c1 =2

and a2 = 2, b2 = k, c2 = 8

Condition for no solution,

1 1 1

2 2 2

a b c

a b c

1 3 2

2 k 8

k = 6

7. Show graphically that each one of the following systems of equations is inconsistent (i.e., has no

solution). [4]

3x – 4y – 1 = 0

82x y 5 0

3

Solution :

We have,

3x – 4 y – 1 = 0

and 8

2x y 5 03



Table for equation 3x – 4y – 1 = 0

3x 1

y4

x 0 3 –1

3x 1y

4

–1/4 2 –1

Points 1

A 0,4

B(3, 2) C(–1, –1)

Now, plot all the points on a graph paper and join them to get a line AB.

Table for equation 2x – 8

3y + 5 = 0

6x – 8y +15 = 0 y =6x 15

8

x 0 –2.5 1.5

6x 15y

8

1.875 0 3

Points D(0, 1875) E(–2.5, 0) F(15, 3)

Now, we plot all the points on a graph paper and join them to get a line EF.

We find the lines represented by equations 3x – 4y – 1 = 0 and 2x –8

3+ 5 = 0 are parallel. So, the

two lines have no common point. Hence, the given system of equations is inconsistent.

8. Solve for x and y

ax + by – a + b = 0

and bx – ay – a – b = 0. [4]

Solution :



Given, ax + by – a + b = 0

and bx – ay – a – b = 0.

The given equations may be written as

ax + by = a – b ...(i)

and bx – ay = a + b ...(ii)

On multiplying Eq. (i) by a and Eq. (ii) by b, we get

a2x + aby = a2 - ab ...(iii)

and b2x – aby = ab + bz ...(iv)

On adding Eqs. (iii) and (iv), we get

(a2 +b2)x = a2 +b2

2

2 2

a b2x 1

a b

On putting x = 1 in Eq. (i), we get

a × 1+ by = a – b

a = by = a – b

by = a – b – a

by = – b

y = – 1

x = 1 and y = – 1 is the required solution.

9. Find the value of k for which the system of equations kx – y =2, 6x – 2y = 3 has (i) a unique solution (ii)

no solution. Is there a value of k for which the given system has infinitely many solutions? [4]

Solution:


kx – y – 2 = 0 and 6x – 2y – 3 = 0.

This is of the form

a1x + b1y + c1 = 0

and a2x + b2y + c2 = 0

where, a1 = k, b1 = – 1, c1 = – 2

and a2 = 6, b2 = –2, c2 = – 3

(i) For a unique solution, we must have

1 1

2 2

a b

a b

k 1

6 2



k 1

6 2

k 3

(ii) For no solution, we must have

1 1 1

2 2 2

a b c

a b c

k 1 2

6 2 3

k 1 2

6 2 3

k 1

6 2 and

k 2

6 3

k = 3 and k 4

Clearly, k = 3 also satisfies the condition k 4.

Hence, the given system will have no solution when k = 3.

Further, the given system will have infinitely many solutions only when

1 1 1

2 2 2

a b c

a b c .

k 1 2

6 2 3

k 1 2

6 2 3

This is never possible.

Since, 1 2

2 3

Hence, there is no value of k for which the given system of equations has infinitely many solutions.

10. Two places A and Bare 100 km apart on a highway. One car starts from A and another from B at the

same time. If the cars travel in the same direction at different speeds, they meet in 5 h. If they travel

towards each other, they meet in 1 h. What are the speeds of the cars? [4]

Solution :

Let the speed of two cars be x km/h and y km/h, respectively.

Case I When two cars travel in the same direction, they will meet each other at P after 5



The distance covered by a car from A = 5x km

[ distance = speed × time]

and distance covered by another car from B = 5y km

5x – 5y = 100

x – y = 2 [divide both sides by 5] ...(i)

Case II When two cars travel in opposite (towards) direction, they will meet each other at Q after 1 h.

The distance covered by a car from A = x km and distance covered by another car from

B = y km


x + y = 100 km ...(ii)

On adding Eqs. (i) and (ii), we get

x y 100

x y 200

2x 120

120

x2

x = 60 km/h

On putting x = 60 in Eq. (ii), we get

60 + y =100 y =100 – 60

y = 40 km/h

Hence, the speed of two cars are 60 km/h and 40 km/h, respectively.

Pioneer Education {The Best Way To Success} IIT JEE ... the following system of linear equations by...

Documents

Transcript of Pioneer Education {The Best Way To Success} IIT JEE ... the following system of linear equations by...