Solving systems of equations using Substitution method and elimination

zee-f · 2011-06-21 05:04:57

i need help solving the following
for the first one this is what i did
or at least try to do
y=(2/3)x-1
y=-x-4
-x+4=(2/3)x-1

1. y = (2/3)x - 1

y = -x + 4

2. x + y = 0
3x + y = -4

3. 4x + 3y = -15
y = x + 2

4. x + 2y = -4
4y = 3x + 12

5. y = 2x
x + y = 3

6. x = 3 - 3y
x + 3y = -6

7. y = -2x + 1
y = x - 5

8. y = (1/2)x - 3
y = (3/2)x 1

9. x + y = 2
4y = -4x + 8
Solve numbers 10 18 using the elimination method. Show your work.
10. y = (2/3)x - 1
y = -x + 4

11. x + y = 0
3x + y = -4

12. 4x + 3y = -15
y = x + 2

13. x + 2y = -4
4y = 3x + 12

14. y = 2x
x + y = 3

15. x = 3 - 3y
x + 3y = -6

16. y = -2x + 1
y = x - 5

17. y = (1/2)x - 3
y = (3/2)x 1

18. x + y = 2
4y = -4x + 8

19. Why are the substitution and elimination methods necessary when we already know how to solve systems of linear equations using the graphing method?

anonimnystefy · 2011-06-21 05:33:05

hi zee-f

As for 19. sometimes it's easier to use these methods instead of graphing.
If in an equation you had 5/7x+3/11, then it would certainly be easier to use one of these to methods.
Just so you know the elimination method is a form of the Gaussian method which can be used not only for systems of two but also three and more equations.

Last edited by anonimnystefy (2011-06-21 05:35:08)

zee-f · 2011-06-21 05:41:05

thanx for the help

zee-f · 2011-06-26 02:56:24

heyyy i solved the following and i was wondering if i did them correct

#5- y=2x
x+y=3

x+2x=3
-2x -2x

x+2x=3
-2x -2x
x+2x=3x

3x=3
x=1
y=2*1

#6- x=3-3y
x+3y=-6

x= 3-3y
3-3y+3y=-6

-3y+3y=0
3+0=-6
3=-6

so it has no solution

#7- y=-2x+1
y=x-5

-2x+1=x-5
-1 -1

-2x=-6+x
-2x=-6+x

-2x+-1x=-6+x+-1x

-2x+-1x=-3x
-3x=-6+x+-1x
x+-1x=0
-3x=-6+0
-3x=-6
x=2

-2*2+1=-3
y=-3

#8-
y= (1 / 2) x-2
y= (3 / 2) x-1

1/2x-2=3/2x-1

added -1/2x to each side

-2+1/2+-3/2

-2+1/2x+-3/2x=-1+3/2x+-3/2

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

#9-
x+y=2
4y=-4x+8

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

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

4y=-4(2+-1y)+8
4y=-8+4y+8

-8+8=0
4y=0+4y
4y=4y

4y+4y=4y+-4y
4y+-4y=0

0=4y+-4y
4y+-4y=0
0=0

bobbym · 2011-06-26 03:02:00

Hi zee-f;

A little easier on 5):

Divde both sides by 3.

Correction pointed out by anonimnystefy.

6) Is correct but you are still not combining like terms to simplify things.

7) x = 2 is correct. You can work it this way:

Add 5 to both sides.

Add 2x to both sides

Divide both sides by 3:

x = 2.

8) Is a bit different. Try multiplying the top equation by 3 instead. It looks like you got x = -1 for an answer which is correct. What about y?

9) Is correct the two equations are the same. There are an infinite number of answers.

zee-f · 2011-06-26 09:20:07

am really confused on how to do the elimination method

bobbym thanx alot for the help

bobbym · 2011-06-26 09:23:57

Hi;

I guess you mean #8?

Elimination is trying to get rid of some variables. Trying to make it easy for you. If you subtract 1) from 2) the y's will disappear. Do you see that?

zee-f · 2011-06-26 09:29:20

like for this one i have to do numbers 10-18 in the elimination method for number
10. y = (2/3)x - 1
y = -x + 4

so am i spouse to subtract the -x and the 2/3 ??

bobbym · 2011-06-26 09:32:27

Hi;

Same as what I said before, if you subtract 1) from 2) the y's will disappear and you will be left with only x.

zee-f · 2011-06-26 09:36:36

really so whats the point of doing the elimination method am ganna end up with x in the end :? am still confused

zee-f · 2011-06-26 09:38:04

10. y = (2/3)x - 1
y = -x + 4

11. x + y = 0
3x + y = -4

12. 4x + 3y = -15
y = x + 2

13. x + 2y = -4
4y = 3x + 12

14. y = 2x
x + y = 3

15. x = 3 - 3y
x + 3y = -6

16.
y = -2x + 1
y = x - 5

17. y = (1/2)x - 3
y = (3/2)x 1

18. x + y = 2
4y = -4x + 8

these are the problems am talking about

like am i ganna end up with x in all of them ???

bobbym · 2011-06-26 09:47:59

Hi;

Let's do it then.

If you subtract the top one ( 1 ) from the bottom one you will get y - y = 0. -x - (2/3)x = -(5/3)x. And 4 - (-1) = 5

So you should get:

Now you have a single variable equation. These are called linear equations. Can you solve that?

zee-f · 2011-06-26 09:54:06

yup am working on it this is ganna take a while am slow in math

zee-f · 2011-06-26 09:55:12

so x is ganna be 3

bobbym · 2011-06-26 10:00:13

Very good! That is correct!

zee-f · 2011-06-26 10:09:18

really thats it thats the elimination method

bobbym · 2011-06-26 10:13:49

Yes, that is all there is to it. Sometimes you will have to multiply an equation by a number to make elimination work. See you a little later.

zee-f · 2011-06-27 01:03:47

i was wondering if i did number 4 correct
x + 2y = -4
4y = 3x + 12

x+2y=-4
x+2y+-2y=-4+-2y
2y+-2y=0
x+0=-4+-2y
x=-4+-2y
4y=3(-4+-2y)+12
4y=(-12+-6y)+12
-12+12=0
4y=0+-6y
4y=-6y
4y+6y=-6y+6y
4y+6y=10y
10y=-6y+6y
10y=0
y=0

anonimnystefy · 2011-06-27 01:26:19

excuse me bobbym but in post #5 you divided the whole equation by 3. so 3/3 is not 3 but 1.

also zee-f try using latex.saved me a lot of time and trouble.

Last edited by anonimnystefy (2011-06-27 01:29:22)

bobbym · 2011-06-27 01:33:26

Hi anonimnystefy;

Thanks for spotting that.

Hi zee-f;

You got the correct answer.

zee-f · 2011-06-27 01:47:08

thanx bobbym and anonimnystefy i will thanx

bobbym · 2011-06-27 01:48:21

Hi zee-f;

Your welcome, you are still doing more work than you have to on some of the problems but you are getting them right.

anonimnystefy · 2011-06-27 01:51:31

hi zee-f

This is how post #18 would look with LaTeX:

Much neater.

Last edited by anonimnystefy (2011-06-27 01:52:05)

zee-f · 2011-06-27 01:54:36

for number 11 i tried to do it using the elimination method
x+y=0
3x+y=-4

4x+y=-4
4x=-4
x=-1

3x-1+y=-4
3x-1
-3+y=-4
-3+3+y=-4+3
-3-3=0
0+y=-4+3
y=-4+3
y=-1

zee-f · 2011-06-27 01:55:38

thank alot anonimnystefy

Math Is Fun Forum

#1 2011-06-21 05:04:57

Solving systems of equations using Substitution method and elimination

#2 2011-06-21 05:33:05

Re: Solving systems of equations using Substitution method and elimination

#3 2011-06-21 05:41:05

Re: Solving systems of equations using Substitution method and elimination

#4 2011-06-26 02:56:24

Re: Solving systems of equations using Substitution method and elimination

#5 2011-06-26 03:02:00

Re: Solving systems of equations using Substitution method and elimination

#6 2011-06-26 09:20:07

Re: Solving systems of equations using Substitution method and elimination

#7 2011-06-26 09:23:57

Re: Solving systems of equations using Substitution method and elimination

#8 2011-06-26 09:29:20

Re: Solving systems of equations using Substitution method and elimination

#9 2011-06-26 09:32:27

Re: Solving systems of equations using Substitution method and elimination

#10 2011-06-26 09:36:36

Re: Solving systems of equations using Substitution method and elimination

#11 2011-06-26 09:38:04

Re: Solving systems of equations using Substitution method and elimination

#12 2011-06-26 09:47:59

Re: Solving systems of equations using Substitution method and elimination

#13 2011-06-26 09:54:06

Re: Solving systems of equations using Substitution method and elimination

#14 2011-06-26 09:55:12

Re: Solving systems of equations using Substitution method and elimination

#15 2011-06-26 10:00:13

Re: Solving systems of equations using Substitution method and elimination

#16 2011-06-26 10:09:18

Re: Solving systems of equations using Substitution method and elimination

#17 2011-06-26 10:13:49

Re: Solving systems of equations using Substitution method and elimination

#18 2011-06-27 01:03:47

Re: Solving systems of equations using Substitution method and elimination

#19 2011-06-27 01:26:19

Re: Solving systems of equations using Substitution method and elimination

#20 2011-06-27 01:33:26

Re: Solving systems of equations using Substitution method and elimination

#21 2011-06-27 01:47:08

Re: Solving systems of equations using Substitution method and elimination

#22 2011-06-27 01:48:21

Re: Solving systems of equations using Substitution method and elimination

#23 2011-06-27 01:51:31

Re: Solving systems of equations using Substitution method and elimination

#24 2011-06-27 01:54:36

Re: Solving systems of equations using Substitution method and elimination

#25 2011-06-27 01:55:38

Re: Solving systems of equations using Substitution method and elimination

Board footer