I cant seem to find the answer to this can anyone help me?

Question

anonymous · Answer

attachment

jdoe0001 · Answer

well, do you know how to do system of equations?

anonymous · Answer

yeahbut i cant seem to find the answer i get (4,6) -.- ? @jdoe0001

jdoe0001 · Answer

$$
\begin{matrix}
	5x &+7y & =32 \
	8x & +6y & =46 \
\hline
	\end{matrix}
	ext{multiply the 2nd by -1}\
\begin{matrix}
	5x &+7y & =32 \
	8x & +6y & =46 & 	imes -1 \
\hline\
? & ? & ? 
	\end{matrix}
$$

luigi0210 · Answer

matrix seemed easier

anonymous · Answer

its the last one... use cramers rules

anonymous · Answer

(5,1)

jdoe0001 · Answer

matrix?   w000t?

jdoe0001 · Answer

hmmm

jdoe0001 · Answer

hold on

jdoe0001 · Answer

-1 will not the trick

anonymous · Answer

thanks @drotzmann

amistre64 · Answer

id suggest just testing out the options to see what fits ....

anonymous · Answer

No problem, i do suggest using youtube to learn how to do Cramer's Rule

anonymous · Answer

omg i had totaly forgor that system @amistre64

anonymous · Answer

youtubes super helpful too i also forgot about that too ._. @drotzmann

anonymous · Answer

haha:)

jdoe0001 · Answer

well, to cancel out the "y" I'd just do 

6 * n = 7
n = 7/6

thus 6 * 7/6 = 7

that gives you a 7y at the bottom, thus cancelling out "y"

amistre64 · Answer

in kindergarten, i was confronted by a hole shaped like a triangle, and 4 blocks oddly shaped blocks.  instead of measuring stuff and trying to work out some complicated things ... i just picked up each block till i found the one that fit :)

jdoe0001 · Answer

$$
\begin{matrix}
	5x &+7y & =32 \
	8x & +6y & =46 \
\hline
	\end{matrix}\
	ext{multiply the 2nd by -7/6}\
\begin{matrix}
	5x &+7y & =32 \
	-\frac{28}{3}x & -7y & =-\frac{161}{3} & 	imes \color{blue}{-\frac{7}{6}} \
\hline\
? & 0 & ? 
	\end{matrix}
$$

anonymous · Answer

@jdoe0001 ....one of us is on crack... i have no idea what you are doing. Cramer's Rule is so much more easier

jdoe0001 · Answer

hehhe, yes, because is a 2x2, well, yeah, I gather you'd use that :)

anonymous · Answer

three 2x2's to be correct ;)

jdoe0001 · Answer

hehe, yes, not sure ... if it'd be .... I guess it depends
I mean, off you have to get the discriminant, then 2 more discriminants

jdoe0001 · Answer

determinants rather

anonymous · Answer

yup:) exactly