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Mathematics 21 Online
OpenStudy (anonymous):

does anyone knows the craimer's rule here??

OpenStudy (amistre64):

for which part?

OpenStudy (anonymous):

first part

OpenStudy (amistre64):

y = na-mc /ad-bc

OpenStudy (anonymous):

a 1 x + b 1 y = c 1 (1) a 2 x + b 2 y = c 2 (2) solve that in craimers rule..

OpenStudy (amistre64):

thats better :)

OpenStudy (anonymous):

can you solve it??

OpenStudy (amistre64):

\[x = \frac{c2b1-c1b2}{a2b1 - a1b2}\] \[y = \frac{c2a1-c1a2}{a1b2-a2b1}\]

OpenStudy (amistre64):

i perfer not to have so many variables the same

OpenStudy (anonymous):

b 2 a 1 x + b 2 b 1 y = b 2 c 1 (1) - b 1 a 2 x - b 1 b 2 y = - b 1 c 2 (2) ____________________________ b 2 a 1 x - b 1 a 2 x = b 2 c 1 - b 1 c 2 i got that by multiply equation (1) by b 2 and equation (2) by -b 1 and add the right and left hand terms

OpenStudy (amistre64):

ax + by = n (1) cx + dy = m (2) \[x = \frac{mb-nd}{cb-ad}\] \[y = \frac{ma-nc}{ad-bc}\]

OpenStudy (amistre64):

you did fine

OpenStudy (anonymous):

tnx..

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