Question

1.)

Answer 1

Answer:

Ques 1)

(-2,12)

Ques 2)

The solution is: (4,1)

Ques 3)

The solution is:

(-4,1)

Ques 4)

The solution is: (-3, 3)

Ques 5)

The solution is: (1,-2)

Step-by-step explanation:

Ques 1)

We have to solve the following system of equation using elimination method.

{5x+y=2

4x+y=4

we will subtract equation (2) from first to obtain:

5x-4x=2-4

x= -2

Now on putting the value of x in first equation we obtain:

5×(-2)+y=2

-10+y=2

y=2+10

y=12

Hence, the solution is:

(-2,12)

Ques 2)

Now again we have to solve using linear combination method.

{2d−e=7

d+e=5

we will add both the equations to get:

2d+d=12

3d=12

d=4

and on putting the value of d in second equation we obtain:

4+e=5

e=5-4

e=1

Hence,

C. The solution is (4, 1).

Ques 3)

5x−y=−21

x+y=−3

we will add both the equation to obtain:

5x+x=-21-3

6x = -24

x= -4

and on putting the value of x in equation (2) we get:

-4+y = -3

y= -3+4

y=1

Hence, the solution is:

(-4,1)

Ques 4)

−3x+9y=36

4x+12y=24

we will divide first equation on both side by 3 and second equation on both side by 4 to obtain the system as:

-x+3y=12

x+3y=6

on adding both the equations we get:

3y+3y=12+6

i.e. 6y=18

i.e. y=3

Hence on putting the value of y in one of the equation we obtain:

x= -3

Hence, the solution is:

(-3,3)

Ques 5)

7/2x−1/2y=9/2

3x−y=5

on multiplying both side of the equation by 2 we obtain:

7x-y=9

Now on subtracting second equation from this transformed equation we obtain:

7x-3x=9-5

4x=4

x=1

Hence on putting the value of x in one of the equations we obtain the value of y as:

y= -2

Hence, the solution is:

(1, -2)