Question

Solve the following by substitution method.

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Answer 1

(3)

The given equations are

• 2x + 3y = 20 .....(i)
• 3x + 4y = 25 .....(ii)

From (i) no. equation, we get

2x = 20 - 3y

⇒ x = (20 - 3y)/2 .....(iii)

Substituting x = (20 - 3y)/2 in (ii) no. equation, we get

3 (20 - 3y)/2 + 4y = 25

⇒ (60 - 9y + 8y)/2 = 25

⇒ 60 - y = 50

⇒ y = 10

Putting y = 10 in (iii), we get

x = {20 - 3 (10)}/2

⇒ x = (20 - 30)/2

⇒ x = - 10/2

⇒ x = - 5

Therefore, the required solution is

x = - 5 , y = 10

(3)

The given equationa are

• 3x + 4y = 0 .....(i)
• x - 5y = 19 .....(ii)

From (ii), we get

x = 5y + 19 .....(iii)

Substituting x = 5y + 19 in (i) no. equation, we get

3 (5y + 19) + 4y = 0

⇒ 15y + 57 + 4y = 0

⇒ 19y = - 57

⇒ y = - 3

Putting y = - 3 in (iii), we get

x = 5 (- 3) + 19

⇒ x = - 15 + 19

⇒ x = 4

Therefore, the required solution is

x = 4 , y = - 3