Solving systems of equations using Substitution method and elimination

bobbym · 2011-06-27 01:57:35

Hi zee-f;

Please recheck your work that answer is not what I am getting.

Subtract the top one from the bottom equation.

anonimnystefy · 2011-06-27 02:06:42

hi zee-f

if you would like to do this using the elimination than it would be like this:

zee-f · 2011-06-27 03:30:25

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 03:31:38

thanx guys

anonimnystefy · 2011-06-27 03:32:21

hi zee-f

look at posts #26 an #27

zee-f · 2011-06-27 05:35:24

am ganna need help doing these in the elimination method
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

anonimnystefy · 2011-06-27 05:44:18

hi zee-f

you should learn when it's better to use substitution method.for example , in 12) its easier just to substitute x from the second equation into the first one.

bobbym · 2011-06-27 05:46:59

Hi zee-f;

14,15,16,17 just subtract the first equation from the second. In each case that will reduce the simultaneous set to a single linear equation.

I believe that we have already done 17 and 18.

zee-f · 2011-06-27 06:42:54

hey anonimnystefy yeah it is really much easier to do it in the substitution method but am asked to do the following for an online algebra course she wants me to show in the elimination method and am finding it really hard

bobbym · 2011-06-27 06:43:47

Did you try what I suggested in the post right above yours?

zee-f · 2011-06-27 06:44:10

bobbym thanx
ooh and i was wondering is this first step correct 4x+3y-x+2=-15

bobbym · 2011-06-27 06:50:36

If that is number 12 then you did not use elimination.

zee-f · 2011-06-27 09:41:09

4x + 3y = -15 + 2 + -1x

Combine like terms: -15 + 2 = -13
4x + 3y = -13 + -1x

4x + 3y = -13 + -1x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add 'x' to each side of the equation.
4x + x + 3y = -13 + -1x + x

Combine like terms: 4x + x = 5x
5x + 3y = -13 + -1x + x

Combine like terms: -1x + x = 0
5x + 3y = -13 + 0
5x + 3y = -13

Add '-3y' to each side of the equation.
5x + 3y + -3y = -13 + -3y

Combine like terms: 3y + -3y = 0
5x + 0 = -13 + -3y
5x = -13 + -3y

Divide each side by '5'.
x = -2.6 + -0.6y

anonimnystefy · 2011-06-27 10:12:06

hi zee-f

What you did there isn't correct.
Elimination method is about eliminating one of the two variables.
So it would be like this:

Last edited by anonimnystefy (2011-06-27 10:13:05)

zee-f · 2011-06-28 01:41:39

thanx alot its just the elimination method is sooo much harder then the substitution
i tried to do this one
in the elimination
x + 2y = -4
4y = 3x + 12

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

4x+14y=0

4x=-14y
x=-3.5y

anonimnystefy · 2011-06-28 01:52:42

hi zee-f

i think your understanding things the wrong way.
first of all you cant say that 12+2y=14y.
second like i said in post#39 you have to get rid of either x or y so that you can calculate the other one.
look again at post #39 lines 5,6,7 in latex.you'll see that the y is gone from one of the equations leaving only the x.then we can calculate x like in a normal equation.

zee-f · 2011-06-28 01:57:22

so should i subtract instead ??

anonimnystefy · 2011-06-28 01:59:13

what you first have to do is multiply the equations so that coefficients with x become the same.

zee-f · 2011-06-28 02:21:36

Add '-2y' to each side of the equation.
x + 2y + -2y = -4 + -2y

Combine like terms: 2y + -2y = 0
x + 0 = -4 + -2y
x = -4 + -2y

Simplifying
x = -4 + -2y

anonimnystefy · 2011-06-28 02:34:20

i'llshow you what i mean in this problem and then try to do the next one just like that.

Last edited by anonimnystefy (2011-06-28 02:35:56)

zee-f · 2011-06-28 03:17:49

ok so like i totaly drag i know but i tried like when i started this is what i did
y=2x
x+y=3

y=2x
x+y=3

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

bobbym · 2011-06-28 03:28:26

Hi zee-f;

How are you today? You did not do any eliminating over there.

anonimnystefy · 2011-06-28 03:29:07

again you need to get rid of y.and you still have y in your equation.

Bob · 2011-06-28 03:35:59

hi zee-f

y=3-x
y=2x

So y is equal to two things. Put the two things equal to each other.

Not

3-x+2x=y

But rather

3-x = 2x

Now the 'y' is 'eliminated'.

Hope that helps

Bob

Last edited by Bob (2011-06-28 03:37:25)

zee-f · 2011-06-28 07:50:10

hey:)

so i did numbers 15-18 i tried to do them this is what i got

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

x=3-3y
-3 -3

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

3-x=3y
x+3y=-6
4x-3y+x=9y
3x-3y=9y

3x=6

x=2

#16- y=-2x+1
y=x-5
y=3x+6
y=3x+6
x=2

y-6=3x

y=-3x
x=2

#17- y=1/2x-3
y=3/2x-1

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

1x--4=y

#18-

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

-4x+8=4y
x+y=2
-3x+8y=6y

-3x=-2y

Math Is Fun Forum

#26 2011-06-27 01:57:35

Re: Solving systems of equations using Substitution method and elimination

#27 2011-06-27 02:06:42

Re: Solving systems of equations using Substitution method and elimination

#28 2011-06-27 03:30:25

Re: Solving systems of equations using Substitution method and elimination

#29 2011-06-27 03:31:38

Re: Solving systems of equations using Substitution method and elimination

#30 2011-06-27 03:32:21

Re: Solving systems of equations using Substitution method and elimination

#31 2011-06-27 05:35:24

Re: Solving systems of equations using Substitution method and elimination

#32 2011-06-27 05:44:18

Re: Solving systems of equations using Substitution method and elimination

#33 2011-06-27 05:46:59

Re: Solving systems of equations using Substitution method and elimination

#34 2011-06-27 06:42:54

Re: Solving systems of equations using Substitution method and elimination

#35 2011-06-27 06:43:47

Re: Solving systems of equations using Substitution method and elimination

#36 2011-06-27 06:44:10

Re: Solving systems of equations using Substitution method and elimination

#37 2011-06-27 06:50:36

Re: Solving systems of equations using Substitution method and elimination

#38 2011-06-27 09:41:09

Re: Solving systems of equations using Substitution method and elimination

#39 2011-06-27 10:12:06

Re: Solving systems of equations using Substitution method and elimination

#40 2011-06-28 01:41:39

Re: Solving systems of equations using Substitution method and elimination

#41 2011-06-28 01:52:42

Re: Solving systems of equations using Substitution method and elimination

#42 2011-06-28 01:57:22

Re: Solving systems of equations using Substitution method and elimination

#43 2011-06-28 01:59:13

Re: Solving systems of equations using Substitution method and elimination

#44 2011-06-28 02:21:36

Re: Solving systems of equations using Substitution method and elimination

#45 2011-06-28 02:34:20

Re: Solving systems of equations using Substitution method and elimination

#46 2011-06-28 03:17:49

Re: Solving systems of equations using Substitution method and elimination

#47 2011-06-28 03:28:26

Re: Solving systems of equations using Substitution method and elimination

#48 2011-06-28 03:29:07

Re: Solving systems of equations using Substitution method and elimination

#49 2011-06-28 03:35:59

Re: Solving systems of equations using Substitution method and elimination

#50 2011-06-28 07:50:10

Re: Solving systems of equations using Substitution method and elimination

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