更新1:
Using substitution :')
How do you solve this simultaneous equation? 4x+y=4 2x+3y=-3?
回答 (5)
2017-02-20 12:41 am
✔ 最佳答案
4x + y = 42x + 3y = -3
5y = -10
y = -2
x = 3/2
2017-02-19 11:43 pm
4x + y = 4 …… [1]
2x + 3y = -3 …… [2]
From [1] :
y = 4 - 4x …… [3]
Substitute [3] into [2] :
2x + 3(4 - 4x) = -3
2x + 12 - 12x = -3
-10x = -15
x = 3/2
Substitute x = (3/2) into [3] :
y = 4 - 4(3/2)
y = -2
Hence, x = 3/2, y = -2
2x + 3y = -3 …… [2]
From [1] :
y = 4 - 4x …… [3]
Substitute [3] into [2] :
2x + 3(4 - 4x) = -3
2x + 12 - 12x = -3
-10x = -15
x = 3/2
Substitute x = (3/2) into [3] :
y = 4 - 4(3/2)
y = -2
Hence, x = 3/2, y = -2
2017-02-19 11:42 pm
The second equation can be written as:
2x = -3 - 3y
i.e. 4x = -6 - 6y
Substituting in to the first equation we have:
(-6 - 6y) + y = 4
i.e. -6 - 5y = 4
=> 5y = -10
so, y = -2
Then, 4x - 2 = 4
=> 4x = 6
i.e. x = 3/2
:)>
2x = -3 - 3y
i.e. 4x = -6 - 6y
Substituting in to the first equation we have:
(-6 - 6y) + y = 4
i.e. -6 - 5y = 4
=> 5y = -10
so, y = -2
Then, 4x - 2 = 4
=> 4x = 6
i.e. x = 3/2
:)>
2017-02-20 1:08 am
-12x - 3y = - 12
2x + 3y = - 3_____add
-10x = - 15
x = 5/2
10 + y = 4
y = - 6
x = 5/2 , y = - 6
2x + 3y = - 3_____add
-10x = - 15
x = 5/2
10 + y = 4
y = - 6
x = 5/2 , y = - 6
2017-02-19 11:50 pm
There is more than one way.
Substitution:
rearrange the first equation to solve for y in terms of x
y = -4x+4
substitute -4x+4 for y in the second equation
2x+3(-4x+4) = -3
solve for x
2x-12x+12 = -3
-10x = -15
x = 3/2
y = -4x+4 = -4(3/2)+4 = -2
:::::
Elimination:
multiply second equation by -2
-4x-6y = 6
add to first equation to eliminate the x terms
4x+y = 4
-4x-6y = 6
------------
-5y = 10
y = -2
Substitution:
rearrange the first equation to solve for y in terms of x
y = -4x+4
substitute -4x+4 for y in the second equation
2x+3(-4x+4) = -3
solve for x
2x-12x+12 = -3
-10x = -15
x = 3/2
y = -4x+4 = -4(3/2)+4 = -2
:::::
Elimination:
multiply second equation by -2
-4x-6y = 6
add to first equation to eliminate the x terms
4x+y = 4
-4x-6y = 6
------------
-5y = 10
y = -2
收錄日期: 2021-04-20 19:35:49
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https://hk.answers.yahoo.com/question/index?qid=20170219153900AAE9yAE