Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

  
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

776,733 users

Find 2x - 6y = -11, 3x - 6y = 11

+3 votes
related to an answer for: How to work out the problem
asked Jan 3, 2013 in ALGEBRA 2 by Johncena Apprentice

3 Answers

+1 vote
2x – 6y = -11   -----(1)

3x – 6y = 11   -----(2)

Multiply the eqation (1)  with  3  and

Multiply the eqation (2)  with   -2

  3×(2x – 6y = -11) =    6x -18y =   -33

 -2×(3x – 6y = 11)  =  -6x +12y = -22

Now add the resultant two eqations.Inorder to add 2 eqations, group like terms and combine them.

6x + (-6x) -18y +12y = -33 -55

⇒  -6y = - 88       

⇒  y = 44/3

Now substitute y= 44/3 in eqation (1)

     2x -6(44/3) =-11

⇒  2x -88 = -11

⇒  2x -88+88 = -11 +88

⇒  2x = 77

⇒  x = 77/2

The values of x and y are  44/3, 77/2  respectively
answered Jan 5, 2013 by peterson Rookie

The values of x and y are 22 , 55/6 respectively.

+2 votes
2x – 6y = -11   -----(1)

3x – 6y = 11   -----(2)

subtract  the  equation 1 & 2

      2x – 6y = -11   -----(1)

(-)   3x – 6y = 11   -----(2)

--------------------------------------

- x              = -22

now subtution eqution  1

2 (22) - 6y = -11

44  - 6y = -11

- 6y  =-11 -44

-6y   = -55

    y = -55/6

The values of x and y are  - 22 , - 55/6 respectively
answered Jan 8, 2013 by krish Pupil

-x = -22 => x = 22.

-6y = -55 => y = 55/6

The values of x and y are 22 , 55/6 respectively.

+3 votes
2x – 6y = -11   -----(1)

3x – 6y = 11   -----(2)

subtract  the  equation 1 & 2

      2x – 6y = -11   -----(1)

(-)   3x – 6y = 11   -----(2)

--------------------------------------

- x              = -22

         x=22

now subtution eqution  1

2 (22) - 6y = -11

44  - 6y = -11

- 6y  =-11 -44

-6y   = -55

    y = 55/6

The values of x and y are   22 , 55/6 respectively
answered Jan 11, 2013 by krish Pupil

Related questions

...