how do you solve 2c-d=-2,4c+d=20?
回答 (6)
2009-07-10 7:33 am
✔ 最佳答案
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
2009-07-10 2:35 pm
4c-2d=-4
4c+d=20
3d=24
d=8
c=3
4c+d=20
3d=24
d=8
c=3
2009-07-10 2:31 pm
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
â´ c = 3, d = 8
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
â´ c = 3, d = 8
2009-07-10 2:30 pm
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.
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.
2009-07-10 2:29 pm
2009-07-10 2:27 pm
c = 6
d = -4
d = -4
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