Wut would these be? Puttng them into y = form:
3x+2y=6
6x+4y= -12
2007-09-28 06:38:10 · 5 answers · asked by Ryan M 2 in Science & Mathematics ➔ Mathematics
Question unclear but will look at the two equations:-
3 x + 2 y = 6
2 y = - 3 x + 6
y = (- 3 /2 ) x + 3
and
4 y = - 6 x - 12
y = (- 6 / 4) x - 3
y = (- 3 / 2 ) x - 3
Hope that this may be of use.
2007-09-28 10:32:17 · answer #1 · answered by Como 7 · 0⤊ 0⤋
3x + 2y = 6
Take the 3x to the "other side, it changes from positive to negative)
2y = 6 - 3x
now divide both sides by 2
y = 3 - (3x)/2
6x + 4y = -12
(same as above take the Xs across to the other side)
4y = -12 - 6x
Again we divide by the number of Ys to get down to one
4y/4 = -12/4 -6x/4
y = -3 -3x/2
2007-09-28 13:45:17 · answer #2 · answered by David F 5 · 0⤊ 0⤋
1. 3x + 2y = 6
transpose 3x to the other side by subtracting it
2y = 6 - 3x
divide both sides by 2 to get y
y = 3 - 3x/2
2. 6x+ 4y = -12
subtract 6x both sides
4y = -12 -6x
divide both sides by 4 to cancel and remain only y
y= -3 - 3x/2
2007-09-28 13:58:05 · answer #3 · answered by mm 2 · 0⤊ 0⤋
You mean you want to solve each of these individually for y?
For the first one, subtract 3x from both sides, then divide by 2.
For the second one, subtract 6x from both sides, then divide by 4.
2007-09-28 13:41:17 · answer #4 · answered by Anonymous · 1⤊ 0⤋
y = (6 - 3x)/2
y = (-12 - 6x)/4
You do that with simple, elementary, algebraic manipulations.
Doug
2007-09-28 13:45:12 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋