Question

Matrix Question: I think I skipped a step but I got it?

Answer 1

Okayy so b =5 which is correct but when I multiplied 1 & 2 I canceled both a and c and I dont know what equation to plug it into because I already got b

Answer 2

My neck is hurting looking at that sideways...

Answer 3

Lol I'm so sorry I can change it

Answer 4

Is this it?
$$
a+b+c=2\\
a+b-c=-8\\
4a+2b+c=-1
$$

Answer 5

No its this:
a+b+c=-10
a-b+c=-8
4a+(-2b)+c=-1

Answer 6

If we add together the first two equation, we get:
\(a+c=-3\)
so: \(c=-3-a\)

Answer 7

substituting that last equation into the first one, we get:
\(b=2-c-a=2-(-3-a)-a=2+3+a-a=5\)

Answer 8

Thats the thing, @Michele_Laino I eliminated a and c in my work by just multiplying 1 by -1 and add it to equation 2, I was wondering if it's even possible to finish from this step.

Answer 9

summarizing: \(c=-3-a,b=5\)

Answer 10

yes, I think so!

Answer 11

Woah, okay let me take a moment to absorb all the math.

Answer 12

Okay @Michele_Laino You make it seem so easy. Also, what would I do when I just got the b=5? What equation would I plug it into??

Answer 13

@Michele_Laino how did you get `a+c = -3` ? When I add the first two equations, I get `2a+2c = -18` which can be turned into `a+c = -9`

Answer 14

you can substitute \(b=5\) into the first two original equations, so you should get a system of two equations with the unknowns \(a,c\) and you are able to solve it for \(a,c\)

Answer 15

@jim_thompson5910 @Michele_Laino Also, the answer to C=-7

Answer 16

when I sum I get: \(2a+2c=-6\) @jim_thompson5910

Answer 17

correct! @Melissa_Something

Answer 18

a+b+c=-10
a-b+c=-8
-----------
2a+0b+2c = -18


2a+2c = -18
a+c = -9

Answer 19

The solutions are {(4, 5, -7)] By the way, I know how I am when I dont get the right answer!

Answer 20

please we have \(a+b+c=2\) @jim_thompson5910

Answer 21

correct! @Melissa_Something

Answer 22

I'm basing what I see when @Melissa_Something said

`No its this: `
`a+b+c=-10`
`a-b+c=-8`
`4a+(-2b)+c=-1`

Answer 23

@jim_thompson5910 Omg I am so sorry the -10 was supposed to be a 2!!

Answer 24

ah that makes more sense now

Answer 25

thanks for your help @jim_thompson5910

Answer 26

@Michele_Laino Um, so the step after I got b=5... Am I wrong? It looks all weird. I used the equation when I multiplied -1 to equation 5.

Answer 27

no, I think no, in genral when I have to solve a linear system, I try to multiply an equation by a factor and the another equation by another factor, in order to eliminate one unknown.
After that I substitute what i find in the remaining equation, it is a normal procedure

Answer 28

general*

Answer 29

Oh okay, thank you I'll try that now :(

Answer 30

ok! :)

Answer 31

Ok Im confused I got -83b+c =40 @Michele_Laino :(

Answer 32

what are your steps?

Answer 33

If I subtratct the second equation from the first one, I get:
\(2b=10\)

Answer 34

\[(-4)-2b+c=-10(-4)\]
which is 8b+(-4)c=40 then added up is
\[6b+(-3)c=-41\]

Answer 35

Which is confusing because -41 / into 6 is -6.833333 :(

Answer 36

from third equation I get:
\(-2b+c=-1-4a=-1-16=-17\)

Answer 37

please wait

Answer 38

It's fine :c It literally takes me 3 hours to do 1 of these -___-

Answer 39

please, write your algebraic system, so i can help you better

Answer 40

From where :( What algebraic system?

Answer 41

I don't understand your steps :(

Answer 42

I'm just gonna skip this, it makes me anxious and after a while evrything looks the same and I cant even tell how I got something :S I dont even understand my steps, maybe science major isnt for me...

Answer 43

please wait, when we are solving a linear system, it is better to combine only two equations at time, not three

Answer 44

please, don't be discouraged, science is for all!

Answer 45

Haha, that's the thing I only combined 2.. :/ It's okay teaching me math is the worst. After all my questions of why or how it just boils down to because math. & I dont understand :(

Answer 46

another advice is:
you have to combine only original equations, namley you have to make all of your operations, only on the original equations

Answer 47

The thing is I only got b=5 and even when I plug it into the 1st equation its \[a+5+c=2\]

Answer 48

ok! now substitute \(b=5\) into the second one, what equation do you get?

Answer 49

\[a-5+c=-1\]

Answer 50

is the second equation like this:
\(a-b+c=-8\)?

Answer 51

Yes, ugh I dont even know where I got the -1

Answer 52

\[a-5+c=-8\]

Answer 53

so both the first and second equation, give the subsequent condition:
\(a+c=-3\), or equivalently \(c=-3-a\)

Answer 54

equations*

Answer 55

I swear you pulled that out of thin air :c

Answer 56

ok! now, please, write the third equation with \(b=5\) and \(c=-3-a\)

Answer 57

what do you get?

Answer 58

Hold on I literally just understood where you got the -3 from and now I'm mad, haha. Ok Lemme do it

Answer 59

Wait the 3rd original equation, right?

Answer 60

yes!

Answer 61

\[4a-2(5)+(-3-a)=-1\] ?

Answer 62

perfect!
and after a simplification, such equation can be rewritten as follows:
\(4a-10-3-a=-1\)

Answer 63

now, please solve that equation for \(a\)

Answer 64

I can't thank you enough :C You're super patient and you actually care about if someone gets it or not. Is there any way to request a qualified helper as you? Or have pay tutor sessions :(

Answer 65

Okie dokie

Answer 66

at the moment I can help you without the necessity to pay

Answer 67

please solve that equation for \(a\)

Answer 68

I think I did it wrong, hold on :( && Yes I was talking about for the future

Answer 69

other students tag me and I help them without payment

Answer 70

OMG guess what, I got it, I got a=4!!!

Answer 71

perfect!

Answer 72

That's very generous of you, much more respect :(

Answer 73

Okay okay, dont tell me I'll do it!! Hahaha

Answer 74

:)
now, please substitute \(a=4,b=5\) into the first original equation

Answer 75

what do you get?

Answer 76

I got -5!!!

Answer 77

when I make those substitution, into your first original equation, I get:
\(4+5+c=2\)
am I right?

Answer 78

substitutions*

Answer 79

Yes, you're right somehow I got it wrong again -___-

Answer 80

I added 4+5 and subtracted from 2....

Answer 81

Correct!

Answer 82

what is \(c\)?

Answer 83

That's whats weird when I did 2-7 I got -5 Hahaha

Answer 84

:) please retry
we have \(c=2-9=...?\)

Answer 85

Oh wow, of course. C=-7.... I put 4+5=7 Hahahaha Idk from where

Answer 86

Oh my gosh I'm finished!!!

Answer 87

yes! Now you know the general method in order to solve a linear system

Answer 88

You're a life saver... I was on the verge of breaking something.. I got so upset I gave myself a headache :(

Answer 89

Thank you so much :( I wish I was as good as you

Answer 90

thanks! :) Yes, one day you will be better than me!

Answer 91

Hahaha most likely not, maybe in another field but not math, have a good night. I wish you much love & light! :)

Answer 92

thanks!! :)