Solve each system of equations. Let z be the parameter.?

1)

x + 2y + 3z = 11 …(a1)

-------------------------x2

2x + 4y + 6z = 22 …(a2)

A1 and A2 are the same equation

2)

2x – y + z = 2 …(b1)

-------------------------x2

4x – 2y + 2z = 4 …(b2)

B1 and B2 are the same equation

3) A1 and B2

x + 2y + 3z = 11

4x – 2y + 2z = 4

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

5x + 5z = 15 --> x = 3 – z

4) A2 and B1

2x + 4y + 6z = 22

2x – y + z = 2

-------------------------(-)

5y + 5z = 20 --> y = 4 - z

Hope this will help. If there is any other thing i can help, dont hesitate to ask.

Regards,

Marcellus Tanjaya

email: bog3lstyle@yahoo.com.sg

marcellus.tanjaya@gmail.com

What you want to do is to bathe up x & y so that they in reality remember upon z. (a million) x + 2y + 3z = 11 (2) 2x - y + z = 2 From (a million) slove x by using transferring something else from left to good (3) x = 11 -2y -3z Use (3) in to modify x in (2) to get the y 2(11 - 2y - 3z) - y + z = 2 => 22 - 5y - 5z = 2 => 20 - 5z = 5y => y = 4 - z (5) (5) in (3) supplies x = 11 - 2(4 - z) - 3z = 3 - z So the reply is x = 3 - z (6) y = 4 - z (7) the position both x & y in reality relies upon on parameter z as requested. to substantiate that the reply is genuine in basic terms insert (6) & (7) in (a million) and (2) to work out that each and every aspect stay equivalent. (6), (7) => (a million) (3 - z) + 2(4 - z) + 3z = 11 -3z + 3z = 11 (6), (7) => (2) 2(3 - z) -(4 - z) + z = 2 -2z +z + z = 2