ML Aggarwal Solutions For Class 9 Maths Chapter 5 Simultaneous Linear Equations

ML Aggarwal Solutions For Class 9 Maths Chapter 5 Simultaneous Linear Equations gives a clear understanding of the concept of simultaneous linear equations. BYJU’S provides you solutions, which are designed by our subject experts. Students can easily download ML Aggarwal Solutions from our website in PDF format for free. This chapter deals with solving for two variables, when two equations are given. Solving different problems helps students to improve time management skills. It also increases their confidence, so that they can be stress-free during exams.

Two linear equations in two variables, taken together are called simultaneous linear equations. In ML Aggarwal solutions For Class 9 Maths Chapter 5, we come across solving the simultaneous linear equations. Sign in BYJU’S to get more practice on these solutions.

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Access answers to ML Aggarwal Solutions for Class 9 Maths Chapter 5 – Simultaneous Linear Equations

Chapter Test

Solve the following simultaneous linear equations (1 to 4):

1.(i) 2x – (¾)y = 3,

5x – 2y = 7

Solution:

ML Aggarwal Sol Class 9 Maths chapter 5-1

8x-3y = 12 …(i)

5x-2y = 7 …(ii)

Multiply (i) by 5 and (ii) by 8, we get

40x-15y = 60 (iii)

40x -16y = 56 (iv)

Subtract (iv) from (iii), we get

y = 4

Substitute y in (i)

8x-3×4 = 12

8x = 12+12

8x = 24

x = 24/8

x = 3

Hence x = 3 and y = 4.

(ii) 2(x-4) = 9y+2

x – 6y = 2

Solution:

2(x-4) = 9y+2

2x-8 = 9y+2

2x-9y = 2+8

2x-9y = 10 ….(i)

x-6y = 2 ….(ii)

Multiply (ii) by 2, we get

2x -12y = 4 ….(iii)

Subtract (iii) from (i), we get

2x-9y = 10

-2x +12y = -4

——————

0+3y = 6

3y = 6

y = 6/3

y = 2

Substitute the value of y in (i)

2x-9×2 = 10

2x-18 = 10

2x = 10+18

2x = 28

x = 28/2

x = 14

Hence x = 14 and y = 2.

2. (i) 97x+53y = 177

53x+97y = 573

Solution:

Given equations are as follows.

97x+53y = 177 …(i)

53x+97y = 573 …(ii)

Multiply (i) by 53 and (ii) by 97

53(97x+53y) = 53×177

5141x+2809y = 9381 …..(iii)

97(53x+97y) = 97×573

5141x+9409y = 55581 …..(iv)

Subtract (iv) from (iii)

5141x+2809y = 9381 …..(iii)

5141x+9409y = 55581 …..(iv)

——————————

0x -6600y = -46200

-6600y = -46200

y = -46200/-6600

y = 7

Substitute the value of y in (i)

97x+53×7 = 177

97x+371 = 177

97x = 177-371

97x = – 194

x = -194/97

x = -2

Hence x = -2 and y = 7.

(ii) x+y = 5.5

x-y = 0.9

Solution:

x+y = 5.5 …(i)

x-y = 0.9 …(ii)

——————–

Adding (i) and (ii), we get

2x = 5.5+0.9

2x = 6.4

x = 6.4/2

x = 3.2

Substitute value of x in (i)

3.2+y = 5.5

y = 5.5-3.2

y = 2.3

Hence x = 3.2 and y = 2.3.

3. (i) x+y = 7xy

2x-3y+xy = 0

Solution:

x+y = 7xy …(i)

2x-3y+xy = 0 …..(ii)

Divide (i) by xy, we get

ML Aggarwal Sol Class 9 Maths chapter 5-2

Divide (ii) by xy, we get

ML Aggarwal Sol Class 9 Maths chapter 5-3

Multiplying (iii) by 3, we get

ML Aggarwal Sol Class 9 Maths chapter 5-4

Adding (v) and (iv), we get

ML Aggarwal Sol Class 9 Maths chapter 5-5

Substitute value of y in (iv)

ML Aggarwal Sol Class 9 Maths chapter 5-6

Hence x = 1/3 and y = 1/4.

(ii)

ML Aggarwal Sol Class 9 Maths chapter 5-7

Solution:

ML Aggarwal Sol Class 9 Maths chapter 5-8

Multiply (i) by 4 and (ii) by 3, we get

ML Aggarwal Sol Class 9 Maths chapter 5-9

Subtracting (iv) from (iii), we get

ML Aggarwal Sol Class 9 Maths chapter 5-10

Substitute (v) in (i), we get

ML Aggarwal Sol Class 9 Maths chapter 5-11

Now solve for (v) and (vi)

x+y = 11

x-y = 5

Add (v) and (vi)

2x = 16

x = 16/2 = 8

Substitute x in (v)

8+y = 11

y = 11-8

y = 3

Hence x = 8 and y = 3.

4. (i) ax+by = a-b

bx-ay = a+b

Solution:

ax+by = a-b …(i)

bx-ay = a+b …(ii)

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

a(ax+by) = a(a-b)

a2x +aby = a2-ab …(iii)

b(bx-ay) = b(a+b)

b2x -aby = ab+b2 …(iv)

Adding (iii) and (iv)

a2x +aby = a2-ab

b2x -aby = ab+b2

———————–

(a2+b2)x = (a2+b2)

x = (a2+b2)/ (a2+b2)

x = 1

Substitute the value of x in (i), we get

a×1+by = a-b

a+by = a-b

by = -b

y = -b/b

y = -1

Hence x = 1 and y = -1.

(ii) 3x + 2y = 2xy

ML Aggarwal Sol Class 9 Maths chapter 5-12

Solution:

3x + 2y = 2xy …(i)

ML Aggarwal Sol Class 9 Maths chapter 5-13

Divide (i) by xy

ML Aggarwal Sol Class 9 Maths chapter 5-14

Multiply (ii) by 2, we get

ML Aggarwal Sol Class 9 Maths chapter 5-15

Subtract (iii) from (iv)

ML Aggarwal Sol Class 9 Maths chapter 5-16

———————-

ML Aggarwal Sol Class 9 Maths chapter 5-17

Substitute y in (iii)

(3/3) + (2/x) = 2

1+(2/x) = 2

(2/x) = 1

x = 2

Hence x = 2 and y = 3.

5. Solve 2x -(3/y) =9

3x + (7/y) = 2.

Hence find the value of k if x = ky + 5.

Solution:

2x -(3/y) = 9 …(i)

3x + (7/y) = 2 …(ii)

Multiply (i) by 3 and (ii) by 2, we get

6x-(9/y) = 27 ..(iii)

6x+(14/y) = 4 …(iv)

Subtracting (iv) from (iii), we get

-23/y = 23

y = 23/-23

y = -1

Substitute y in (i)

2x-(3/-1) = 9

2x+3 = 9

2x = 9-3

2x = 6

x = 6/2

x = 3

Hence x = 3 and y = -1.

Given x = ky+5

Substitute x and y in above eqn

3 = k×-1+5

3 = -k+5

k = 5-3

k = 2

Hence the value of k is 2.

6.

ML Aggarwal Sol Class 9 Maths chapter 5-18

Hence find the value of 2x2-y2.

Solution:

ML Aggarwal Sol Class 9 Maths chapter 5-19

Let (x+y) = a

(1/a)-(1/2x)

ML Aggarwal Sol Class 9 Maths chapter 5-20

Multiply (iii) by 5

ML Aggarwal Sol Class 9 Maths chapter 5-21

Subtracting (ii) from (iv)

ML Aggarwal Sol Class 9 Maths chapter 5-22

Substitute x in (iii)

(1/a) -1/(2×3) = 1/30

(1/a) – (1/6) = 1/30

1/a = (1/30)+(1/6)

1/a = (1+5)/30

1/a = 6/30

a = 30/6

a = 5

Substitute a in x+y = a

3+y = 5

y = 5-3

y = 2

Hence x = 3, y = 2.

2x2-y2 = 2×32-22

= 2×9-4

= 18-4

= 14

Hence the value of 2x2-y2 is 14.

7. Can x, y be found to satisfy the following equations simultaneously ?

ML Aggarwal Sol Class 9 Maths chapter 5-23

If so the find.

Solution:

ML Aggarwal Sol Class 9 Maths chapter 5-24

Multiply (i) by 5 and (ii) by 2, we get

ML Aggarwal Sol Class 9 Maths chapter 5-25

Subtract (v) from (iv)

31/x = 95-2

31/x = 93

x = 31/93

x = 1/3

Substitute x in (i)

(2/y)+5÷(1/3) = 19

(2/y)+5×3 = 19

(2/y) = 19-15

(2/y) = 4

y = 2/4

y = 1/2

Substitute x and y in (iii)

3×(1/3) + 8×(1/2) = 5

1+4 = 5

The value of x and y satisfies (iii).

Hence the given equations are simultaneous.

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