solving by elimination
1)
2x-y=3
4x+3y=21
when i solve it i get x=1.2 and y= -.6 but when i check it back in to the equation it is wrong will someone help me!

- anonymous

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- jim_thompson5910

2x-y=3
4x+3y=21
3*(2x-y)=3*3
4x+3y=21
6x-3y=9
4x+3y=21
-----------
10x = 30
10x = 30
x = 30/10
x = 3
Now that we know that x = 3, we can use it to find y
2x-y = 3
2(3)-y = 3
6 - y = 3
-y = 3-6
-y = -3
y = 3
So x = 3 and y = 3

- anonymous

thank you so much! would you be willing to explain this one?
x-y=2
-2x+2y=5

- jim_thompson5910

x-y=2
-2x+2y=5
2(x-y)=2*2
-2x+2y=5
2x-2y=4
-2x+2y=5
-----------
0x+0y = 9
0 = 9 ... which is FALSE
So there are no solutions. This means that the system is inconsistent.

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## More answers

- anonymous

okay the zero threw me off when i had done it! do you know how to solve by using back substitution by chance?

- jim_thompson5910

Solving by substitution:
x-y=2
x = 2 + y
------------------------------------------
-2x+2y = 5
-2(2+y) + 2y = 5
-4 - 2y + 2y = 5
-4 = 5 ... which is FALSE
So again, the system is inconsistent and there are no solutions.

- anonymous

here is one...
2x-y+5z=24
y+2z=4
z=6

- anonymous

for using back-substitution

- jim_thompson5910

y+2z=4
y+2(6)=4
y + 12 = 4
y = 4-12
y = -8
-------------------------------------------------------
2x-y+5z = 24
2x - (-8) + 5(6) = 24
2x + 8 + 30 = 24
2x + 38 = 24
2x = 24-38
2x = -14
x = -14/2
x = -7
So the solutions are x = -7, y = -8 and z = 6

- anonymous

you are incredible thank you!

- jim_thompson5910

You're welcome

- anonymous

hey are you still on here? can you help me with solve using row-operations...
x+y+z=6
2x-y+z=3
3x -z=0

- jim_thompson5910

x+y+z=6
2x-y+z=3
3x -z=0
1 1 1 6
2 -1 1 3
3 0 -1 0
1 1 1 6
2 -1 1 3
0 -3 -4 -18 R3 - 3R1
1 1 1 6
0 -3 -1 -9 R2 - 2R1
0 -3 -4 -18
1 1 1 6
0 1 1/3 3 (-1/3)*R2
0 -3 -4 -18
1 1 1 6
0 1 1/3 3
0 0 -3 -9 R3 + 3R2
1 1 1 6
0 1 1/3 3
0 0 1 3 (-1/3)*R3
x+y+z = 6
y+(1/3)z = 3
z = 3
=============================================
=============================================
y+(1/3)z = 3
y+(1/3)(3) = 3
y + 1 = 3
y = 3-1
y = 2
----------------
x+y+z = 6
x+2+3 = 6
x + 5 = 6
x = 1
So the solutions are x = 1, y = 2 and z = 3

- anonymous

May i ask you what the R stands for

- jim_thompson5910

R stands for the row
So R1 means "row 1", R2 means "row 2", and R3 means "row 3"

- anonymous

okay so how did you determine subtract row 3- 3 by row1

- jim_thompson5910

When going from
1 1 1 6
2 -1 1 3
3 0 -1 0
to
1 1 1 6
2 -1 1 3
0 -3 -4 -18 R3 - 3R1
I did this to make that lower left corner value go from 3 to 0
Notice that 3 - 3*1 = 3-3 = 0
This basically "eliminates" the x variable in equation 3 which gets me one step closer to solving for x, y, and z

- anonymous

okay can you tell me what you get for this one then...
6y+4z=-18
3x+3y = 9
2x -z =12

- jim_thompson5910

using the same method of row reduction or some other method?

- anonymous

row reduction :)

- jim_thompson5910

alright thanks, one sec

- jim_thompson5910

6y+4z=-18
3x+3y = 9
2x -z =12
0 6 4 -18
3 3 0 9
2 0 -1 12
0 6 4 -18
1 1 0 3 (1/3)*R2
2 0 -1 12
1 1 0 3 R1 <--> R2
0 6 4 -18
2 0 -1 12
1 1 0 3
0 6 4 -18
0 -2 -1 6 R3 - 2*R1
1 1 0 3
0 -2 -1 6 R2 <--> R3
0 6 4 -18
1 1 0 3
0 1 1/2 -3 (-1/2)*R2
0 6 4 -18
1 1 0 3
0 1 1/2 -3
0 0 1 0 R3 - 6*R2
x+y = 3
y + (1/2)z = -3
z = 0
-------------------------------------------------------
y + (1/2)z = -3
y + (1/2)(0) = -3
y = -3
--------------------
x+y = 3
x + (-3) = 3
x - 3 = 3
x = 3+3
x = 6
So the solutions are x = 6, y = -3, and z = 0

- anonymous

THANK YOU! You don't know how much you have helped me! there is only one more I can't seem to get right do you have time for one more? :)

- jim_thompson5910

sure whats the problem

- anonymous

same thing so solving by row operations :)... its
x-2y+5z=2
4x -z=0

- jim_thompson5910

x-2y+5z=2
4x -z=0
1 -2 5 2
4 0 -1 0
1 -2 5 2
0 8 -21 -8 R2 - 4*R1
x-2y+5z = 2
8y-21z=-8
-------------------------------------------------------
8y-21z=-8
8y = -8 + 21z
y = (21z - 8)/8
-----------------------
x-2y+5z = 2
x-2( (21z - 8)/8 )+5z = 2
x - 4(21z - 8) + 5z = 2
x - 84z + 32 + 5z = 2
x - 79z + 32 = 2
x - 79z = 2-32
x - 79z = -30
x = 79z - 30
Now if you let z = s, where s is any number, then the solutions are
x = 79s - 30, y = (21s - 8)/8, z = s
This means that there are an infinite number of solutions.
So the system is consistent and dependent.

- jim_thompson5910

A very quick way to determine if you'll have a dependent system is noticing that there are more variables than equations.

- anonymous

so the answer is consistent and dependent or x = 79z - 30
sorry I am a little confused haha

- jim_thompson5910

Well both aspects make up the total answer.
The idea here is that there are an infinite number of points where the two equations x-2y+5z=2 and 4x -z=0 intersect
So this makes the system dependent (since one equation technically "depends" on the other -- it changes as the other one does)
Because at least one solution exists, this makes the system consistent.
The answers of x = 79s-30, etc... are the more explicit forms of the solutions and are more specific than saying "there are an infinite number of solutions"
This is because you can't just say that something random like (1,2,3) is a solution even though I did say "there are an infinite number of solutions". All of the solutions fit a very specific algebraic form.

- anonymous

That makes sense! Thank you again! I appreciate all your help!

- jim_thompson5910

You're welcome, glad it's all clicking.

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