x/2 + y/3 = 1
3/1(3x/2 + y) = 3 3 in y/3 cancels
2/1(3x + 2y) = 6 2 in 3x/2 cancels
3x + 2y = 6
3x +2y - 6 = 0 ?
2007-09-11 07:34:37 · 7 answers · asked by Matthew K 2 in Science & Mathematics ➔ Mathematics
just checking to see if the movies i did were legal.
2007-09-11 07:45:57 · update #1
Your last two steps are right on. They are equivalent to your original equation.
You do however have a notation problem in the first two steps. Step one, since you chose to multiply both sides by 3/1, should be written:
3/1(x/2 + y/3) = 3/1(1)
This simplifies to:
3x/2 + y = 3
Multiplied by 2 we get:
2/1(3x/2 + y ) = 2/1(3)
This simplifies to:
3x + 2y = 6
I should mention that you might have multiplied both sides by 6/1, The LCD and you would have saved a step.
Good luck, you are on the right track!
2007-09-11 07:50:14 · answer #1 · answered by Peter m 5 · 0⤊ 0⤋
x/2 + y/3 = 1 to get rid of the fractions we need to multiply by 2 and by 3 ....so why not multiply once by 6?
3x+2y=6
so as you rightly say 3x+2y-6=0
2007-09-11 14:50:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Yep.
Further:
2y = -3x +6
y = -3/2 x + 3
2007-09-11 14:39:31 · answer #3 · answered by gebobs 6 · 0⤊ 0⤋
yes, although you confused me with the way you did that.
Why not just multiply by 6 to begin with to get the 3x + 2y = 6?
2007-09-11 14:40:33 · answer #4 · answered by Becky M 4 · 0⤊ 0⤋
Yes, that is done correctly. What you've done is multply the initial equations through by 6.
2007-09-11 14:40:52 · answer #5 · answered by Anonymous · 0⤊ 0⤋
your question is not understandable.
what you are looking for ?
what is given and what is expected ?
I would suggest to post it again clearly mentioning the question.
thanks
2007-09-11 14:39:43 · answer #6 · answered by calculus 1 · 0⤊ 2⤋
no!
2007-09-11 14:40:32 · answer #7 · answered by Dragon'sFire 6 · 1⤊ 1⤋