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Solve the following system of equations: x − 2y = 14 x + 3y = 9 (1, 12) (−1, −12) (12, −1) (12, 1)
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subtract 2nd equation from 1st equation. tell me what u get.
\[\large{x-2y=14}\] \[\large{x+3y=9}\] solve for x in the first equation
it cant be a or d
x = 14+2y now put this value in second one: \[\large{(14+2y)+3y=9}\]
thats correct @Carol329
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Can you solve for y now? @Carol329
its c :)
correct
wow! how do u figure that out ?
good work @Carol329 .
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