# Simultaneous Linear Equation – Exercise 3.5.3 – Class IX

1. Solve the following system of equations by the method of substitution:

(i) x + y = 10

x – y = 12

(ii)3x – 7y = 7

11x + 5y = 87

(iii) 3x – 4y = 10

4x + 3y = 5

(iv) 3a – 2b = 12

4a – 5b = 16

(v) 5x + 4y – 4 = 0

x – 20 = 12y

(vi) 7x + 5y = 10

3x + y = 2

(vii) 2p + 3q = -5

3p – 2q = 12

(viii) 2x + 3y = 5

3x + 4y = 6

Solution

(i) x + y = 10…………….(1)

x – y = 12 ……………..(2)

from (1), we have, x = 10 – y …………..(3)

Substitute (3) in (2), we get,

10 – y – y = 12

10 – 2y = 12

-2y = 12 – 10 = 2

y = –2/2 = -1

Since x = 10 – y , we have y = – 1,

Therefore, x = 10 – (-1) = 10 + 1 = 11

Hence, x = 11 and y = -1

(ii)3x – 7y = 7 ………………….(1)

11x + 5y = 87……………………(2)

From (1), x = 7+7y/3……………….(3)

Substitute (3) in (2), we get,

11(7+7y/3) + 5y = 87

3[11(7+7y/3) + 5y = 87]

11(7 + 7y) + 15y = 261

77 + 77y + 15y = 261

+92y = 261 – 77 = 184

92 y = 184

y = 2.

Since, x = 7+7y/3  , we have y = 2, then,

x = 7+7(2)/3 = 7+14/3 = 21/3 = 9

Thus, x = 9 and y = 2.

(iii) 3x – 4y = 10……………(1)

4x + 3y = 5…………….(2)

From (1), x = 10+4y/3 …………..(3)

Substitute (3) in (2), we get

4(10+4y/) + 3y = 5

3[4(10+4y/) + 3y = 5]

4(10 + 4y) + 9y = 15

40 + 16y + 9y = 15

40 + 25y = 15

25y = 15 – 40 = -25

y = -1

Since, x = 10+4y/ and we have y = -1, then

x = 10+4(-1)/3= 10-4/3 = 6/3 = 2

Therefore, x = 2 and y = -1

(iv) 3a – 2b = 12…………….(1)

4a – 5b = 16………………..(2)

From (1), we have a  = 12 + 2b/3 ………..(3)

Substitute (3) in (2), we get

4(12 + 2b/3) – 5b = 16

3[4(12 + 2b/3) – 5b = 16]

4(12+2b) – 15b = 48

48 + 8b – 15b = 48

-7b = 0

b= 0

Since a = 12 + 2b/3 and b = 0

Then, a = 12+0/3 = 12/3 = 4

Therefore , a = 4 and b = 0

(v) 5x + 4y – 4 = 0⇒5x + 4y = 4….……..(1)

x – 20 = 12y⇒x – 12y = 20………..(2)

From (1), we have, x = 4-4y/5 …………..(3)

Substitute (3) in (2)

4-4y/5 – 20 = 12y

(4-4y/5 – 20 = 12y)x5

4 – 4y – 100 = 60y

-96 = 60y + 4y = 64y

y = -96/64 = -3/2

Since, y = –3/2 then x = 4-4y/5 = [4 – 4(-3/2)]/5 = 4 + 6/5 = 10/5 = 2

Therefore, x = 2 and y = -3/2

(vi) 7x + 5y = 10 …………..(1)

3x + y = 2………………….(2)

From (1)

x = 10-5y/7 …………………..(3)

Substitute (3) in (2)

3(10-5y/7 ) + y = 2

[3(10-5y/7 ) + y = 2]x7

= 3(10 – 5y) + 7y = 14

30 – 15y + 7y = 14

-8y = 14 – 30 = – 16

8y = 16

y = 2

Since y = 2 then x = 10-5(2)/7 = 10-10/7 = 0

(vii) 2p + 3q = -5……………..(1)

3p – 2q = 12………………….(2)

From (1),  p = -5-3q/2 …………(3)

Substitute (3) in (2)

3(-5-3q/2 ) – 2q = 12

[3(-5-3q/2 ) – 2q = 12]x2

3(-5-3q) – 4q = 24

-15 – 9q – 4q = 24

-13q = 24 + 15 = 39

q = 39/-13 = -3

Since q = -3 we have, p = -5-3(-3)/2 = -5+9/2 = 4/2 = 2

Therefore q = -3 and p = 2

(viii) 2x + 3y = 5 ……………….(1)

3x + 4y = 6 ……………………(2)

From (1),  x = 5-3y/2 ……………(3)

Substitute (3) in (2), we get,

3(5-3y/2 ) + 4y = 6

[3(5-3y/2 ) + 4y = 6]x2

3(5 – 3y) + 8y = 12

15 – 9y + 8y = 12

-y = 12 – 15 = -3

y = 3

Since y = 3, we have x = 5-3y/2 = 5-3(3)/2 = 5-9/2 = –4/2 = -2

Therefore, y = 3 and x = -2

1. Use elimination method for solving the following equations:

(i) 3x + 2y = 0

x + 3y = 7

Solution:

(3x + 2y = 0)x1⇒ 3x + 2y = 0

(x + 3y = 7)x3 ⇒ 3x + 9y = 27

(-)    (-)    (-)

—————————-

-7y = -27

y = 3

Substituting y = 3 in equation (1)

3x + 2×3 = 0

3x + 6 = 0

3x = -6

x = -2

Therefore, x = -2 and y = 3

(ii) x + y = 3

3x – y = 5

Solution:

x + y = 3……………..(1)

3x – y = 5……………(2)

(1)x3 ⇒ 3x + 3y = 9

(2)x1 ⇒ 3x – y = 5

(-)      (+)     (-)

—————————-

4y = 4

y = 1

Substitute the value of y in  eq(1)

x + y = 3

x = 3 – 1 = 2

Therefore, x = 2 and y = 1

(iii) x –  y = 7

3x + y = 10

Solution:

x –  y = 7 …………………(1)

3x + y = 10……………….(2)

1)x3 ⇒ 3x – 3y = 21

(2)x1 ⇒ 3x + y = 10

(-)      (-)      (-)

——————————–

-4y = 11

y = –11/4

Substitute the value of y in (1)

x –  y = 7

x – (-11/4) = 7

x = 7 – 11/4 = 28-11/4 = 17/4

Therefore, x = 17/4 and y = -11/4

(iii) 3x + y = 10

x–y = 2

Solution:

3x + y = 10 ………………….(1)

x–y = 2…………………………(2)

1)x1 ⇒ 3x + y = 10

(2)x3 ⇒ 3x – 3y = 6

(-)      (+)     (-)

——————————–

4y = 4

y =1

Substitute the value of y in (2)

x–1 = 2

x = 2 + 1 = 3

Therefore x = 3 and y = 1

(v) 2x + y = 0

3x – y = -5

Solution:

2x + y = 0  ………………..(1)

3x – y = -5 ………………..(2)

(1)x3 ⇒ 6x + 3y = 0

(2)x2 ⇒ 6x – 2y = – 10

(-)      (+)     (+)

———————————–

5y = 10

y =2

Substitute the value of y in (1)

2x + y = 0

2x + 2 = 0

2x = – 2

x = -1

Therefore, x = -1 and y = 2

(vi) 2x + y = 7

2x – 3y = 3

Solution:

2x + y = 7 ……………(1)

2x – 3y = 3……………(2)

(-)  (+)   (-)

———————–
4y = 4

y = 1

Substitute the value of y in (1)

2x + y = 7

2x + 1 = 7

2x = 7 – 1 = 6

x = 3

Therefore, x = 3 and y = 1

(vii) 100x + 200y = 700

200x + 100y = 800

Solution:

100x + 200y = 700………….(1)

200x + 100y = 800………….(2)

(1)x2 ⇒ 200x + 400y = 1400

(2)x1 ⇒ 200x + 100y = 800

(-)      (-)                (-)

————————————-
300y = 600

y = 600/300 = 2

Substitute the value of y in (1)

100x + 200(2) = 700

100x = 700 – 400 = 300

x = 300/100 = 3

Therefore, x = 3 and y = 2

(viii) 41x + 53y = 135

53x + 41y = 147

Solution:

41x + 53y = 135……………(1)

53x + 41y = 147……………(2)

Adding (1) and (2), we get

94x + 94y = 282 ⇒ x + y = 3 ………….(3)

Subtracting (1) and (2), we get

41x + 53y = 135

53x + 41y = 147

(-)      (-)      (-)

———————-

-12x +12y = -12 ⇒ -x + y = -1 …………………(4)

From (3) and (4)

x + y = 3

-x + y = -1

—————–
2y = 2

y = 1

Substitite the value of y in (3) i.e,

x + 1= 3

x = 3 – 1 = 2

Therefore, x = 2 and y = 1

1. Solve the following system by evaluating proportional value of the variables:

(i) 5x – 4y = -14

3x + 2y = -4

Solution:

(5x – 4y = -14) x 4  ⇒ 20x – 16y = -56

(3x + 2y = -4) x 14 ⇒42x + 28y = -56

(-)      (-)       (+)

————————————————–
-22x – 44y = 0

-x – 2y=0
x = -2y

Therefore,

x/-2 = y/1 = k

i.e., x = -2k and y = k

substituting in (1)

3(-2k) + 2k = -4

-6k + 2k = -4

-4k = -4

k = -4/-4 = 1

x = -2(1) = -2

y = k = 1

Therefore, x = -2 and y = 1

(ii) 3x + 2y = 5

5x – 4y = 23

Solution:

3x + 2y = 5———(1)

5x – 4y = 23———–(2)

(3x + 2y = 5)x23⇒69x + 46y = 115

(5x – 4y = 23)x5 ⇒ 25x – 20y = 115

(-)      (+)     (-)

————————————————-
44x + 66y = 0

2x + 3y = 0

2x = -3y

Therefore, x/-3­ = y/2 = k

x = -3k and y = 2k————(*)

Then substitute x and y in (1)

3(-3k) + 2(2k) = 5

-9k + 4k = 5

-5k = 5

k = -1

Substitute k in (*)

x = -3(-1)=3 and y = 2(-1) = -2

Therefore, x = 3 and y = -2

(iii) 2x + 3y = 13

4x + y = 11

Solution:

2x + 3y = 13………………….(1)

4x + y = 11…………………….(2)

(2x + 3y = 13)x11⇒22x + 33y = 143

(4x + y = 11)x13 ⇒52x + 13y = 143

(-)       (-)         (-)

———————————————–
-30x + 20y = 0

-3x + 2y = 0

3x = 2y

x/2 = y/3 = k

x = 2k and y = 3k

Substitute the value of x and y in (1)

2(2k) + 3(3k) = 13

4k +9k = 13

13k = 13

k = 13/13 = 1

Then, x = 2(1) = 2 and y = 3(1) = 3

Therefore, x = 2 and y = 3

(iv)3x + 2y = 1

2x + 3y = 4

Solution:

3x + 2y = 1…………….(1)

2x + 3y = 4……………..(2)

(3x + 2y = 1)x4⇒12x + 8y = 4

(2x + 3y = 4)x1⇒2x + 3y = 4

(-) (-)     (-)

—————————————–
10x + 5y = 0

2x + y = 0

2x = -y

Thus, x/-1 = y/2 = k

x = -k and y = 2k

Substitute x and y in (1)

3x + 2y = 1

3(-k) + 2(2k) = 1

-3k + 4k = 1

k = 1

Hence, x = -k = -1

and

y = 2k = 2(1) = 2

Therefore, x = -1 and y = 2

(v) 3x + 2y = 4

2x – 3y = 7

Solution:

3x + 2y = 4……………..(1)

2x – 3y = 7……………….(2)

(3x + 2y = 4)x7⇒21x + 14y = 28

(2x – 3y = 7)x4⇒8x – 12y = 28

(-)   (+)    (-)

—————————————————–
13x+26y=0

x + 2y = 0

x = -2y

Thus x/-2 = y/1 = k

⇒ x = -2k and y = k

Substitute x and y in equation (1),we get

3x + 2y = 4

3(-2k) + 2(k) = 4

-6k + 2k = 4

-4k = 4

k = -1

Hence, x = -2(-1) = 2

y = k = -1

Therefore, x = 2 and y = -1

(vi) 2x + 5y = -3

5x + 2y = -18

Solution:

2x + 5y = -3………………..(1)

5x + 2y = -18………………(2)

(2x + 5y = -3)x6⇒12x + 30y = -18

(5x + 2y = -18)x1⇒5x + 2y = -18

(-)    (-)     (+)

———————————————-
7x + 28y = 0

x + 4y = 0

x = -4y

Thus, x/-4 = y/1 = k

⇒ x = -4k and y = k

Substitute x and y in equation(1)

2x + 5y = -3

2(-4k) + 5k = -3

-8k + 5k = -3

-3k = -3

k = 1

Hence, x = -4 and y = 1

(xii) 3x + y = 7

x – y = 5

Solution:

3x + y = 7……………..(1)

x – y = 5………………..(2)

(3x + y = 7)x5 ⇒15x +5y = 35

(x – y = 5)x7    ⇒7x – 7y = 35

(-)   (+)   (-)

—————————————–
8x + 12y = 0

2x + 3y = 0

2x = -3y

Thus, x/-3 = y/2 = k

⇒ x = -3k and y = 2k

Substitute x and y in eq(2)

x – y = 5

-3k – 2k = 5

-5k = 5

k = -1

Hence, x = -3(-1) = 3

y = 2(-1) = -2

Therefore, x = 3 and y = -2

(vii) 2x – y = 6

3x + y = 9

Solution:

2x – y = 6…………………(1)

3x + y = 9…………………(2)

(2x – y = 6)x3⇒6x – 3y = 18

(3x + y = 9)x2⇒ 6x + 2y = 18

(-)   (-)     (-)

——————————————-
0.x – 5y = 0

Thus, x/-5 = y/0 = k

⇒ x = -5k and y = 0

Substitute x and y in eq(1)

2x – y = 6

2(-5k) – 0 = 6

-10k – 0 = 6

-10k = 6

k =- 6/10 = –3/5 .

Hence, y = -5k = -5(-3/5) = 3

and x = 0.

Therefore, x = 0 and y = 3