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Mathematics 16 Online
OpenStudy (anonymous):

how do i solve these equations using cramers rule? https://indva.brainhoney.com/Resource/935377,0,0,0,0/Assets/flvs/educator_algebra2_v8_gs/Module4/ImagMod4/m4_12_29.gif

OpenStudy (anonymous):

what??

OpenStudy (helder_edwin):

sorry

OpenStudy (turingtest):

I don't think you can make fractions with latex matrices like that here @helder_edwin

OpenStudy (anonymous):

still lost...

OpenStudy (turingtest):

\[\large x=\frac{\left|\begin{matrix}-3&7\\-1&3\end{matrix}\right|}{\left|\begin{matrix}5&7\\2&3\end{matrix} \right|}\]

OpenStudy (turingtest):

haha I win :)

OpenStudy (helder_edwin):

sorry again.for x you have \[ \large \left|\begin{array}{cc} -3 & 7\\ -1 & 3 \end{array}\right| \] over \[ \large \begin{vmatrix} 5 & 7\\ 2 & 3 \end{vmatrix} \]

OpenStudy (turingtest):

do you have any idea where those numbers are coming from @katiebugg ?

OpenStudy (anonymous):

ok.... thanks i think and no

OpenStudy (turingtest):

do you know how to find the augmented matrix of a system like\[ax+by=p\]\[cx+dy=q\]?

OpenStudy (anonymous):

no

OpenStudy (turingtest):

just take the coefficients from the LHS and make them into a matrix with the x coefficients in one column and the y's coefficients in the next the first row has the first equation and the second has the second's example: from the form\[ax+by=p\]\[cx+dy=q\]the augmented matrix is\[A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\]

OpenStudy (turingtest):

let me know if you understood that or where it confuses you

OpenStudy (turingtest):

@katiebugg please do not leave if you are still there, that would be quite rude...

OpenStudy (anonymous):

i didnt leave lol i got it now sorry

OpenStudy (turingtest):

cool, so do you know how to take the determinant of that matrix?

OpenStudy (turingtest):

\[ax+by=p\]\[cx+dy=q\]the augmented matrix is\[A=\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\]the determinant is written\[\det A=\left|\begin{matrix}a&b\\c&d\end{matrix}\right|=?\]do you know how to take that determinant?

OpenStudy (turingtest):

@katiebugg it is hard to help you when you don't reply, and is rather of frustrating to those trying

OpenStudy (anonymous):

i was taking a test sorry i understand it now i looked it upp sorry

OpenStudy (turingtest):

you mean you are done with this question?

OpenStudy (anonymous):

no i skiped the lesson

OpenStudy (turingtest):

so are we going to continue here or not?

OpenStudy (anonymous):

u dont have too

OpenStudy (turingtest):

ok I'll close it then

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