Question

Solve the following pair of linear equation by the elimination method and sublimation method 1. X+y = 5 and 2x -3y = 4 2.3x+ 4y =10 and 2x -2y =2
3. 3x- 5y = 0 and 9x= 2y + 7

Answer 1

Elimination Method

1) x+y=5................ (1)

2x-3y=4............(2)

Multiplying (1) by 3 we get,

3x+3y=15..........(3)

Adding (2) and (3) we get,

5x=19.

x=19/5

Substituting the value of x in (1) we get,

19/5+y=5.....(1)

y=5-19/5

y=25-19/5

y=6/5

Substitution Method

x+y=5........ (1)

2x-3y=4..........(2)

From (1)

x=5-y

Now, Substituting the value of x in (2) we get,

2(5-y) -3y=4

10-2y-3y=4

10-5y=4..

-5y=4-10

-5y=-6

y=-6/-5

y=6/5

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

x+6/5=5 ..... (1)

x=5-6/5

x=25-6/5..

x=19/5

Substitution Method

(2) 3x+4y=10............(1)

2x-2y=2

I. e. x-y=1............. (2)

From (2)

x=1+y

Substituting the value of x in (1) we get,

3(1+y) +4y=10....... (1)

3+3y+4y=10

3+7y=10

7y=10-3

7y=7..

y=7/7

y=1

. Substituting the value of y in (2) we get,

x-1=1....... (2)

x=1+1

x=2

Elimination method

3x+4y=10..........(1)

2x-2y

I. e x-y=1.......... (2)

Multiplying eq (2( by 4 w get,

4x-4y=4.....(3)

Adding (1) and (3)

7x=14

x=14/7

x=2

Substituting the value of x in (2)

2-y=1

-y=1-2

-y=-1

y=1

Elimination method

(3) 3x-5y=0.............(1)

9x=2y+7

I. e. 9x-2y=7...........(2)

Multiplying (1) by 2 and (2) by 5 we get,

6x-10y=0...........(3)

45x-10y=35..........(4)

Subtracting (3) from (4) we get,

39x=35

x=35/39

Substituting the value of x in (1) we get,

3(35/39) -5y=0...... (1)

35/13-5y=0

-5y=-35/13

y= -35/13×1/-5

y= -35/13×-1/5

y=7/13

Substitution method

3x-5y=0.......(1)

9x=2y+7

I. e. 9x-2y=7.........(2)

From (1)

3x-5y=0

3x=5y

x=5y/3

Substituting the value of x in (2) we get,

9(5y/3) -2y=7....... (2)

15y-2y=7

13y=7

y=7/13

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

3x-5(7/13) =0

3x-35/13=0

3x=35/13

x=35/13×1/3

x=35/39

Hope it will help you!

#Siddhi