Question

WILL GIVE BRAINLIEST
Consider the following system:
Equation 1: Ax+By=C
Equation 2: Dx+Ey=F

A,B,C,D,E, and F are non zero real numbers. Which of the following could replace equation 1 and still have the same solution? Select all that apply.

Answer 1

The answers are
Number 1-A multiple of equation 1
Number 2-The sum of equation 1 and 2
Number 5-The sum of the multiple of equation 1 and 2

Answer 2

Answer with explanation:

The given system of equation is

Equation 1: Ax+By=C

Equation 2: D x+E y=F

let,replace, A,B,C , D,E and F,by some integral value, and then we will check, which options are true among options provided.

1.x+ y=2

2.x-y=4

Adding (1) and (2)

2 x=6

x=3 (Dividing both side by,3)

Substituting the value of , x in equation 2,we get

3-y=4

y=-1

So, x=3,and y=-1, is the solution of the system of equation.

Option 1:

Multiply equation 1, by any number,suppose A

1.Ax + A y=2 A

2. x - y =4

To,solve it we have to multiply equation 2, by A also.

2.A x - A y = 4 A →→→Multiplying equation 2, by A

Adding equation (1) and (2)

2 A x= 6 A

x=3

Substitute , x in any of two equation,to get, y= -1.

Option 2:

Adding equation 1, and equation 2, gives

2 x=6

and, x -y=4

Solving ,the above two,yield, x=3, y= -1.

Option 3 and Option 4:

If you replace coefficient of x,and y, in both equation 1,and equation 2, will not yield the same outcome, that is , x=3, and ,y=-1.We can check it by, replacing coefficient of x or y,by any  real number.

Option 5:

Multiply equation 1, by 2 and multiply equation,2 by 3 and add them ,you will get

→2 x + 2 y + 3 x - 3 y=4 +12

1.→5 x - y =16

And ,2.→ x+y =2 or3.→ x-y =4,to obtain value of x and y.

you will find that, x=3, and y=-1 are solution of all the three equation.

→Correct options in the form of statements are

1,2 and 5