Question

(1)x - 3y - 3 = 0 ; 3x - 9y - 2 = 0

(2) 2x + y = 5 ; 3x + 2y = 8

(3) 3x - 5y = 20 ; 6x - 10y = 40

(4) x - 3y - 7 = 0 ; 3x - 3y - 15 = 0

by elimination method only please solve it please. ​

Answer 1

Answer:

1) 4

2) x= 2, y= 1

3)0=0

4)x=4, y=-1

Answer 2

Answer:

#1

solution,

x - 3y = 3 ---- eqn 1

3x - 9y = 2 -------- eqn 2

We have to solve by elimination method

Multiply eqn 1 by -3

-3x + 9y = -9 ----- eqn 3

yo

Add eqn 2 and eqn 3

3x - 9y = 2

-3x + 9y = -9

( + ) -------------

0 = 2 - 9

0 \neq -7

Thus the given system of equations has no solution.

#2

solution,

2x + y = 5 .....eqn(I)

3x + 2y = 8.....eqn(II)

Multiplying eqn(I) by 2

we get

4x + 2y = 5....eqn(III)

subracting eqn II from eqn(III)

4x + 2y = 5

3x + 2y = 8

- - -

x = -3

substituting value of x in eqn I

2x + y = 5

or, 2×(-3) + y = 5

or, -6 + y = 5

so y = 11