Question

Solve each system of equations. Show work/steps on how you got your answer.

19. 4a - 2b + 8c = 30
a + 2b - 7c = -12
2a - b + 4c = 15

Answer 1

Well, basically, the only thing I can tell is to either use substitution or learn matrices. The steps for this are quite long and tedious to display on this site.

Answer 2

Well I am doing elimination and Idk if I need to get rid of the a or b

Answer 3

It doesn't matter what you get rid of first. The procedures are still the same.

Answer 4

@Hero How would you enter these problems into the website???: http://www.wolframalpha.com/

Answer 5

You would enter it as a matrix. Do you know how to translate it into a matrix on paper? Because if you can, I can provide you with a resource.

Answer 6

kindof

Answer 7

Either you know how to do it or you don't. There is no in-between.

Answer 8

give me the website @Hero

Answer 9

http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

Answer 10

Okay I entered them into the martrix. How will I determine what the answer is

Answer 11

look at the last matrix after pressing enter or calculate. The matrix should be in order:

a = 1st row
b = 2nd row
c = 3rd row

Make sure you have entered each number correctly

Answer 12

how many rows and columns should there be I put 3 rows and columns

Answer 13

Yes, three rows and three columns is correct

Answer 14

@SmoothMath I am doing elimination

Answer 15

Pick your favorite equation of the 3 and your favorite variable in that equation. Solve for that variable.
Now that you have that variable solved for, plug it into one of the OTHER 2 equations, which should give you an equation with only 2 variables in it. Pick your favorite variable in that equation and solve for it.
Now that you have that variable solved for, plug into the third equation so that you end up with an equation with only one variable in it.

Answer 16

Can I show you what I have???

Answer 17

Sure.

Answer 18

@smoothmath, what happens if you hate all three equations equally?

Answer 19

4a -2b + 8c = 30
a + 2b - 7c = -12

5a + c = 18

4a - 2b + 8c = 30
-4a + 2b - 8c = -30
0 = 0

Answer 20

Okay, so that's not actually substitution method exactly. You added two equations to each other, which is also a viable method. The first two you added, that does something for you because you got an equation with only 2 variables.
The second equations you added together, that doesn't do anything for you because you're stuck with something totally unhelpful. 0=0? I already knew that.

Answer 21

that's why, because elimination is easier for me

Answer 22

Elimination works nicely as your first step, but after that, there's no nice way to make it work.

Answer 23

You should really learn substitution. It's more versatile. It works more often.

Answer 24

How it would look like, using substitution???

Answer 25

EQ1) x + y + z = 4
EQ2) x - 2y - z = 1
EQ3) 2x - y - 2z = -1

Okay, so I solve EQ1 for x.
x + y + z = 4
x = 4-y-z

Now, I use that to substitute in for x in EQ2.
x - 2y - z = 1
(4-y-z) - 2y -z = 1 {{group like terms}}
4-3y -2z = 1 {{solve for y}}
-3y = =-3 + 2z
y = 1-(2/3)z

Now I use EQ3 and substitute in for x, then for y.
2x -y -2z = -1
2(4-y-z) -y - 2z = -1 {{distribute and simplify}}
8 - 2y - 2z -y -2z = -1
8-3y -4z = -1 {{substitute in for y}}
8-3(1-(2/3)z)-4z = -1 {{distribute and simplify}}
8-3 +2z -4z =-1
5 -2z = -1
-2z = -6
z=3

Now that you know z=3, substitute into previous equations to solve for the other variables.
y = 1-(2/3)z
y = 1-(2/3)(3)
y = 1-2
y = -1

Since you know z=3 and y=-1, substitute into a previous equation to solve for x.
x = 4-y-z
x = 4-(-1)-(3)
x = 5-3
x=2

So, x=2, y=-1, z=3

Answer 26

thank you @SmoothMath

Answer 27

Do you understand? Any questions for me? Think you can do the same process with your problem?

Answer 28

uh @SmoothMath

Answer 29

Oh, you know what, I think that this is one with multiple solutions.

Answer 30

yes it's infinitely many solutions

Answer 31

So in general, if you notice that one of the equations is just a multiple of one of the other equations, then the system is not solvable.

You need 1 equation for each variable to be able to solve the system. In this case, we have 3 variables, so we need 3 equations.

We have 3, but notice that if you double the first equation, it gives us the third equation. That means that the first and the third are essentially the same equation, and so we only have 2 equations for 3 variable, which makes this unsolvable.

Answer 32

How does the 1st equation doubled, give us the 3rd

Answer 33

I apologize. The 3rd doubled gives us the first.

Answer 34

2a - b +4c = 15
2(2a -b + 4c) = 2(15)
4a -2b +8c = 30

Answer 35

so no work to be shown

Answer 36

Hm?

Answer 37

Oh, you want to know how to show that it's unsolvable?

Answer 38

yea

Answer 39

Okay, so if you go through subsitution method correctly, there are 3 different kinds of result you can end up with at the end.
The first is just something like
x=, y=, z=
Meaning you know the exact values that x y and z must take. That's the one and only solution.

The second kind of result you may get is something like
5=5
10=10
0=0
That sort of thing. Some equation which is just necessarily true and not very interesting. If this is your result, then your system has infinite solutions, and the set of solutions will either be a plane or a line. If you were able to get some kind of equation that only had 2 variables in it, then that is a line, and that line is your set of solutions.

The third kind of result you may get is something like
1=0
2=10
-1=1
That sort of thing. Some equation which is just obviously false. That means your system has no solutions.

4a - 2b + 8c = 30
a + 2b - 7c = -12
2a - b + 4c = 15

-2b = 30-4a -8c
b = -15 +2a +4c

a +2(-15+2a+4c)-7c = -12
a -30 +4a+8c -7c = -12
5a +c =18
c = 18-5a

2a -(-15+2a+4c) +4(18-5a) = 15
2a +15-2a-4c +72 -20a = 15
87-4c -20a = 15
87-4(18-5a)-20a = 15
87-72 +20a-20a = 15
15 = 15
So in this case, we have our second kind of solution, infinitely many solutions, and the set of solutions is the line
5a+c = 18

Answer 40

If you have a system of 3 equations with 3 variables, and each equation represents a plane. The set of solutions will be where all 3 planes intersect, which can happen in 3 ways