(2x - 1)(3x +4) = (6x - 2) (x+2)
x equals positive integers
2007-11-26 09:50:28 · 6 answers · asked by dani m 1 in Science & Mathematics ➔ Mathematics
Multiply out on the left using FOIL:
6x² + 8x - 3x - 4 = (6x - 2)(x + 2)
6x² + 5x - 4 = (6x - 2)(x + 2)
Do the same thing on the right side (FOIL):
6x² + 5x - 4 = 6x² + 12x - 2x - 4
6x² + 5x - 4 = 6x² + 10x - 4
Now cancel terms like 6x² and - 4 from both sides:
5x = 10x
Subtract 5x from both sides:
0 = 10x - 5x
0 = 5x
Divide both sides by 5:
0 = x
x = 0
If x has to be a positive integer then there is no solution.
2007-11-26 09:58:09 · answer #1 · answered by Puzzling 7 · 0⤊ 1⤋
X= 0
2007-11-26 10:08:00 · answer #2 · answered by Donga 1 · 0⤊ 0⤋
1) 6x^2+5x-4=6x^2+10x-4
2) move everything to the left(reverse sign when moving past an equal sign). Get 6x^2-6x^2+5x-10x-4+4=0
3) -5x=0
4) x=0
2007-11-26 10:00:47 · answer #3 · answered by Ed S 4 · 0⤊ 0⤋
I can.
(6X² + 8X - 3X - 4) = (6X² +12X - 2X - 4)
6X² + 5X - 4 = 6X² + 10X - 4
-5X = 0
X = 0
(One answerer said "No solution." There is a solution. That solution is zero. Don't forget: zero is still a number!)
2007-11-26 09:57:31 · answer #4 · answered by math geek 3 · 0⤊ 0⤋
6x^2 + 8x - 3x -4 = 6x^2 + 12x - 2x - 4
6x^2 + 5x - 4 = 6x^2 + 10x - 4
5x - 4 = 10x - 4
-4 = 5x - 4
0 = 5x
x = 0
2007-11-26 09:56:43 · answer #5 · answered by Ms. Exxclusive 5 · 0⤊ 1⤋
6x2 + 8x - 3x - 4 = 6x2 + 12x - 2x - 4
5x - 4 = 10x - 4
5x = 10x
no solution
2007-11-26 09:57:13 · answer #6 · answered by xobrigitox 2 · 0⤊ 1⤋