 # 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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Chapter Test

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

1.(i) 2x – (¾)y = 3,

5x – 2y = 7

Solution: 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 Divide (ii) by xy, we get Multiplying (iii) by 3, we get Adding (v) and (iv), we get Substitute value of y in (iv) Hence x = 1/3 and y = 1/4.

(ii) Solution: Multiply (i) by 4 and (ii) by 3, we get Subtracting (iv) from (iii), we get Substitute (v) in (i), we get Now solve for (v) and (vi)

x+y = 11

x-y = 5

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)

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 Solution:

3x + 2y = 2xy …(i) Divide (i) by xy Multiply (ii) by 2, we get Subtract (iii) from (iv) ———————- 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. Hence find the value of 2x2-y2.

Solution: Let (x+y) = a

(1/a)-(1/2x) Multiply (iii) by 5 Subtracting (ii) from (iv) 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 ? If so the find.

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