Question

Given the system of equations:

2x – y = –2
x = 14 + 2y

What is the value of the system determinant?
What is the value of the y−determinant?
What is the value of the x−determinant?
What is the solution to the system of equations?

Answer 1

d ithink

Answer 2

each one is a equation i know its a lot of work... im asking for the system determent the y determent and the x determent plus the solution so ya

Answer 3

@sourwing i know how to do the matrix and for the y determent i have -2 and 28 but what do i do know?! whats the actual formula

Answer 4

@a1234

Answer 5

@wio i know how to do the matrix and for the y determent i have -2 and 28 but what do i do know?! whats the actual formula

Answer 6

@csscgirl same

Answer 7

first, i need a reminder, how do you find the determinant? :P sorry

Answer 8

lol thats what i am trying to figure out but all i know is you set it up like a matrix and do the cross multiplication but after that i am not sure what to do so with the y determinant i did all the steps and am let with 28 and -2 what do i do with them

Answer 9

lol... oh... but the solution to the system is (-6,-10)

Answer 10

i think i might be able to figure it out now, lemme c :P gimme a minute

Answer 11

k thanks

Answer 12

did u use the matrix that is:

2 -1 -2
1 -2 14

Answer 13

I believe if you have the determinants, then \(y = D_y/D\), where \(D_y\) is the \(y\) determinant and \(D\) is the normal determinant.

Answer 14

As per Kramer's rule.

Answer 15

@torchriswil What are you determinants?

Answer 16

Looks like they are not here...

Answer 17

oh sorry and yes i was looking for the formulas

Answer 18

Do you know how to find the determinants or need an explanation?

Answer 19

explanation or formula i dont even know...

Answer 20

i need explanation

Answer 21

Okay, first for the normal determinant. It's simple for a \(2\times 2\) matrix. For larger ones you should look it up yourself.

First of all, this is our equation in matrix form.\[
\begin{bmatrix}
2&-1\\
1&-2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
-2\\
14
\end{bmatrix}
\]

Answer 22

Next for the \(2\times 2\) matrix, our determinant is: \[
\begin{vmatrix}
a&b\\
c&d
\end{vmatrix}
= a\cdot d - b\cdot c
\]

Answer 23

thank you, ill look at this tomorrow, but parents are making me go to bed... it already 10!

Answer 24

Ok, good night then.

Answer 25

thank you so much goodnight :)

Answer 26

this makes it eaier but i still dont under stand... there is different determenants for y x and normal correct?

Answer 27

i mean formulas for the determinants

Answer 28

For the \(x\) determinant, we swap the \(x\) row with the answer row.\[
\begin{bmatrix}
\color{red}2&-1\\
\color{red}1&-2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=
\begin{bmatrix}
\color{blue}{-2}\\
\color{blue}{14}
\end{bmatrix}
\]And then find the determinant: \[
D_x=
\begin{vmatrix}
-2&-1\\
14&-2
\end{vmatrix}
\]

Answer 29

Does this make sense?

Answer 30

yes thank you you explain it way better than my teacher

Answer 31

Tell me what you get.

Answer 32

alright

Answer 33

i have too go to bed but ill get back to you on the answer tomorrow promise

Answer 34

Sure thing, bye.

Answer 35

okay so when you are on can you tell me if i did this part right

Answer 36

a. 2 -1 -1-4= -5
1 2
is this right @wio well for the first one is it?

Answer 37

\[
\begin{vmatrix}
2&-1\\
1&2
\end{vmatrix}
=
(2)(2)-(1)(-1) = 4+1 = 5
\]

Answer 38

ya my friend told me that say 6- (-1) would be positve becuase a double negative equals a positive so the answers would be
a.5
b.-30
c.-26
did i get them right @wio

Answer 39

@wio d i corrected it did i get them right