lets say i had a problem like ...
(3x+6)/(12x+1) = 5(2x+3)/ (x^2+6x+5)
how would i solve this, cross multiply ?
2007-09-06 07:22:37 · 4 answers · asked by dreamz 4 in Science & Mathematics ➔ Mathematics
Thanks everyone
2007-09-06 07:41:09 · update #1
This is a bit more involved but you can, yes:
Distribute 5:
(3x+6)/(12x+1) = (10x+15)/(x^2+6x+5)
Cross multiply:
(3x+6)(x^2+6x+5) = (10x+15)(12x+1)
Left side:
3x^3 + 18x^2 + 15x + 6x^2 + 36x + 30 =
combine like terms:
3x^3 + 24x^2 + 51x + 30
Right side:
120x^2+10x+180x+15
combine like terms:
120x^2+190x+15
Now you have:
3x^3 + 24x^2 + 51x + 30 = 120x^2+190x+15
combine like terms:
3x^3 - 96x^2 - 139x + 15 = 0.
2007-09-06 07:34:10 · answer #1 · answered by miggitymaggz 5 · 1⤊ 0⤋
(3x+6)/(12x+1) = 5(2x+3)/ (x^2+6x+5)
(3x+6)(x^2+6x+5) = (10x+15)(12x+1)
3x^3 + 18x^2 + 15x + 6x^2 +36x + 30 = 120x^2 + 180x + 10x + 15
3x^3 + 24x^2 + 51x + 30 = 120x^2 + 190x + 15
3x^3 + 24x^2 - 120x^2 + 51x - 190x + 30 - 15 = 0
3x^3 - 96x^2 - 139x + 15 = 0
2007-09-06 14:40:08 · answer #2 · answered by robertonereo 4 · 0⤊ 0⤋
yeah probably try and simplify anything if possible (factor out the x^2+6x+5) and try to cross out some terms.
Then cross multiply and solve for x.
2007-09-06 14:36:33 · answer #3 · answered by Anonymous · 1⤊ 0⤋
That's a start. As far as I can tell, you can't cancel any factors, so you're going to have a four-term polynomial. Have fun with all that.
2007-09-06 14:33:11 · answer #4 · answered by gebobs 6 · 0⤊ 0⤋