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replace x in the second equation by y - 3, and get an equation with y alone \[6(y-3)-3y=-9\]
then solve for y. don't forget the distributive law. \[6y-18-3y=-9\] \[3y-18=-9\] \[3y=9\] \[y=3\]
I Got 6y-18-3y=9. . Now What Can I Do ?
Ohhh. Okay. Thank You Sooo Much!
oops i made a mistake. it was y- 6 not y - 3
i goofed. \[6(y-6)-3y=-9\]
\[6y-36-3y=-9\] \[3y-36=-9\] \[3y=27\] \[y=9\]
now since \[x=y-6\] and we know \[y=27\] we get \[x=27-6=21\]
sorry about my typo
thank you, and no problem, thanks!♥
Wait. . But The Multiple Choice Says: A.(-2,-3) B.(-2-2.7) C.(-3,-2) D.(3,5) And 21 Isn't Either of those. . . .
let me check see if i made a mistake
Okay, And It Says To Eliminate, Not Substitute.
ok yes i made a mistake. i wrote \[y=9\] but then for some reason wrote \[x=27-6\] instead of \[x=9-6=3\]
my fault sorry. i do not know why i put y = 9 and then used y = 27 however, i am sticking with (3,9) and the check is easy: first equation x= y - 6 and 3 = 9 - 6 second equation is \[6x-3y=-9\] and \[6\times 3 - 3 \times 9 = 18-27=-9\]
so (3, 9) is right even if it is not one of the choices.
elimination: rewrite \[x=y-6\] as \[x-y=-6\] and put \[6x-3y=-9\] underneath so it looks like \[x-y=-6\] \[6x-3y=-9\]
if you want to eliminate the "y" multiply the top equation all the way across by -3 to get \[-3x+3y=18\] \[6x-3y=-9\]
that way when you add the "y"s will drop out to give \[3x=9\] and so \[x=3\]
dont forget to multiply EVERYTHING by the constant, otherwise you will not have the same equation.
now you know \[x=3\] you can substitute back to find y = 9. as before.
Thank You!! (=