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Default Math - 05-24-2013, 11:11 AM

Solve Linear Systems by Substitution...

Question:

3x+2y-1=0
y=-x+3

2nd Question:

x+4y=5
x+2y=7

Basically I need to end up with Right Side = Left Side
and 2 coordinations like (-1, +5)

Can someone please explain how to do this, I need to complete a chapter by Monday and it requires me knowing how to do this, I've spent a lot of time by myself, but am still stuck.


EDIT: Grade 10 Applied math btw
Ty Ty :P



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Last edited by Revenge2nite; 05-24-2013 at 11:25 AM.
  
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Default 05-24-2013, 11:15 AM

3x+2y-1=0
3x=2y-1
x=(2/3y)-(1/3)

You want the x's? Actually, the same story counts if you want to isolate the y.




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Default 05-24-2013, 11:24 AM

Okay so the teacher did 1 question for me and this is how it came out...


3x+2y-1=0
y=-x+3

3x+2(-x+3)-1=0
3x-2x+6-1=0
1x+5=0
x=-5

then

y=-1x+3
y=-1(-5)+3
y=5+3
y=8

Therefor the coords are (-5, +8)

but the 2nd question I gave it a little different because of the 2nd equation it gives.

I need to find out what x and y are.
@RTFM



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Last edited by Revenge2nite; 05-24-2013 at 11:29 AM.
  
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Default 05-24-2013, 11:25 AM

Got it.

First one. Substitute the "y =" formula into the y in "3x+2y-1=0" Then distribute them.

3x+2y-1=0
y=-x+3


3x+2(-x+3)=0
3x-2x+6=0
x+6=0
x=-6

Second

x+4y=5
x+2y=7

Put one into "Y =" (I chose the bottom one)

x+2y=7
2y=-x+7
y=-x+3.5

then.. substitute like I did in the first one.

x+4(-x+3.5)= 5
x-4x+14= 5
3x+14= 5
3x= -9
x= -3


  
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Default 05-24-2013, 11:27 AM

Quote:
Originally Posted by GodzSkiller View Post
Got it.

First one. Substitute the "y =" formula into the y in "3x+2y-1=0" Then distribute them.

3x+2y-1=0
y=-x+3


3x+2(-x+3)=0
3x-2x+6=0
x+6=0
x=-6

Second

x+4y=5
x+2y=7

Put one into "Y =" (I chose the bottom one)

x+2y=7
2y=-x+7
y=-x+3.5

then.. substitute like I did in the first one.

x+4(-x+3.5)= 5
x-4x+14= 5
3x+14= 5
3x= -9
x= -3
Showed it to the teacher, says it's wrong O_o



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Default 05-24-2013, 11:29 AM

Quote:
Originally Posted by Revenge2nite View Post


Showed it to the teacher, says it's wrong O_o
Really? Did you plug the x back in to get your y?


  
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Default 05-24-2013, 11:30 AM

Quote:
Originally Posted by GodzSkiller View Post


Really? Did you plug the x back in to get your y?
mhm and da fuq is plug the x back in to get your y oh lol the teacher did it for meh... even though it might be wrong i gotta do what the teacher wants me to do, cba doing the right thing and getting marked wrong rather do what the teacher wants :P

are you sure you didn't mess up anywhere?



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Default 05-24-2013, 11:36 AM

Quote:
Originally Posted by Revenge2nite View Post


mhm and da fuq is plug the x back in to get your y oh lol the teacher did it for meh... even though it might be wrong i gotta do what the teacher wants me to do, cba doing the right thing and getting marked wrong rather do what the teacher wants :P

are you sure you didn't mess up anywhere?
I don't think so... except I didn't solve for "y". That is what I was taught.


  
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Default 05-24-2013, 11:38 AM

x+4y=5
x+2y=7

this is one way to do so:

x=5-4y
(5-4y)+2y=7

x=5-4y
5-4y+2y=7

x=5-4y
5-2y=7

x=5-4y
-2y=7-5

x=5-4y
-2y=2

x=5-4.(-1)
y=-1

x=5+4
y=-1

(9,-1)


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Default 05-24-2013, 11:39 AM

Quote:
Originally Posted by GodzSkiller View Post


I don't think so... except I didn't solve for "y". That is what I was taught.
Same, for me.
  
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