Question

x+y=20, 3x+7y=108
..

by elimination method​

Answer 1

Answer:

x+y=20

mana:y=x to

x+x=20

2x=20

x=20÷2=10

x=10

3x+7y=108

x-7y=108÷3

-7y=36x

x=36÷7y

Step-by-step explanation:

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Answer 2

Answer:

Given :-

x + y = 20 and 3x + 7y = 108 by elimination method

Solution :-

x + y = 20 . . . . (1)

3x + 7y = 108 . . . . (2)

Multiply equation (1) by 3, we get

3x + 3y = 60 . . . . (3)

3x + 7y = 108 . . . . (4)

Adding (3) and (4) we get,

3x + 3y = 60

3x + 7y = 108

(-) (-) (-)

____________

- 4y = - 48

y = - 48/- 4

y = 12

Substituting the value of y = 12 in equation (1) we get,

x + y = 20

➜ x + (12) = 20

➜ x + 12 = 20

➜ x = 20 - 12

➜ x = 8

∴ The value of x = 8, y = 12 is the solution of the system of the given equations.