how do you solve 2c-d=-2,4c+d=20?

By substitution ,

2c-d= -2 => 2c= d-2 => c = (d-2)/2 .... (1)
4c+d=20..... (2)

Substituting value of c in equation 2 , we get

4[(d-2)/2] +d =20
=> 2d-4 +d = 20
=> 3d -4 =20
=> 3d = 24
=> d = 8

From equation 1 ,
C= (d-2)/2 = (8-2)/2 = 6/2 =3

So, d=8 , c= 3

4c-2d=-4
4c+d=20
3d=24
d=8
c=3

2c - d = -2
4c + d = 20

2c - d = -2
d = 2c - (-2)
d = (2c + 2) ...... (1)

4c + d = 20

Substitute (1) for d.
4c + (2c + 2) = 20
4c + 2c + 2 = 20
6c = 20 - 2
c = 18/6
c = 3

2c - d = -2

Substitute 3 for c.
2(3) - d = -2
6 - d = -2
d = 6 - (-2)
d = 8

∴ c = 3, d = 8

2c-d=-2 ------> EQN1
4c+d=20 ------>EQN2

adding both, ==> 6c = 18 or c=3

substituting c=3 in eqn1 ==> 6 - d = -2 ==> d=8

c=3, d=8 is the answer.

http://answers.yahoo.com/question/index?qid=20090709231956AAgROoT

c = 6
d = -4