how do you solve 3x-y=9, 2x+y=6?
回答 (4)
2008-09-12 12:43 am
✔ 最佳答案
3x - y = 9; solve for y first. -y = 9 - 3x then y = -9 + 3x or y = 3x - 9.Now that you have that, plug in 3x - 9 into the other equation where y is. it should look like this: 2x + (3x - 9) = 6
5x - 9 = 6
x = 15/5 = 3
now, use that as x and plug back into the first equation to get y:
3(3) - y = 9
y = 0
2016-03-29 2:55 pm
For the best answers, search on this site https://shorturl.im/OPwHe
3x - y = 9 2x + y = 6 Solve for Y in first equation... 3x - y = 9 3x - 9 = y Substitute for Y in second equation... 2x + y = 6 2x + 3x - 9 = 6 5x = 15 x = 3 Substitute for X in first equation... 3x - y = 9 (3*3) - y = 9 9 - y = 9 y = 0 Answer is (3,0)
3x - y = 9 2x + y = 6 Solve for Y in first equation... 3x - y = 9 3x - 9 = y Substitute for Y in second equation... 2x + y = 6 2x + 3x - 9 = 6 5x = 15 x = 3 Substitute for X in first equation... 3x - y = 9 (3*3) - y = 9 9 - y = 9 y = 0 Answer is (3,0)
2015-08-19 3:32 am
This Site Might Help You.
RE:
how do you solve 3x-y=9, 2x+y=6?
RE:
how do you solve 3x-y=9, 2x+y=6?
參考: solve 3x 9 2x 6: https://shortly.im/l0kXm
2008-09-12 12:49 am
3x-y=9 2x+y=6
(3*x)-y=9 (2*x)+y=6
(3/3)=1x-y=(9/3)=x-y= 3 (2/2)*x+y=(6/2)
(x-y= 3) (x+y=3)
(3*x)-y=9 (2*x)+y=6
(3/3)=1x-y=(9/3)=x-y= 3 (2/2)*x+y=(6/2)
(x-y= 3) (x+y=3)
2008-09-12 12:42 am
3x - y = 9 (exchange y & 9 position, when they're on the side, they become the opposite value like -y to +y)
hence y= 3x -9
2x+y = 6
2x + (3x -9) = 6 (substitute y's value from above here)
5x -9 = 6
5x = 6+9
5x = 15
x = 3
this is really an easy one.
hence y= 3x -9
2x+y = 6
2x + (3x -9) = 6 (substitute y's value from above here)
5x -9 = 6
5x = 6+9
5x = 15
x = 3
this is really an easy one.
2008-09-12 12:40 am
so add the equations together to get rid of the y's
3x-y=9
2x+y=6
________________
5x=15
x=3
then substitute in x=3
3x3-y=9
9-y=9
y=0
6+y=6
y=0
3x-y=9
2x+y=6
________________
5x=15
x=3
then substitute in x=3
3x3-y=9
9-y=9
y=0
6+y=6
y=0
收錄日期: 2021-04-29 20:08:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080911163547AAAIlSr