Question

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!

Answer 1

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

Answer 2

thank you so much! would you be willing to explain this one?

x-y=2
-2x+2y=5

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

for using back-substitution

Answer 8

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

Answer 9

you are incredible thank you!

Answer 10

You're welcome

Answer 11

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

Answer 12

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

Answer 13

May i ask you what the R stands for

Answer 14

R stands for the row


So R1 means "row 1", R2 means "row 2", and R3 means "row 3"

Answer 15

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

Answer 16

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

Answer 17

okay can you tell me what you get for this one then...

6y+4z=-18
3x+3y = 9
2x -z =12

Answer 18

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

Answer 19

row reduction :)

Answer 20

alright thanks, one sec

Answer 21

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

Answer 22

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? :)

Answer 23

sure whats the problem

Answer 24

same thing so solving by row operations :)... its

x-2y+5z=2
4x -z=0

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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