[(5x+2)/(x^2-81)] - [(4x-7)/(x^2-81)]
Please help! any help would be appreciated!
2007-12-13 14:13:36 · 6 answers · asked by alex 1 in Science & Mathematics ➔ Mathematics
[(5x+2)/(x^2-81)] - [(4x-7)/(x^2-81)]
The two terms already have a common denominator...
[(5x+2) - (4x-7)] / (x^2 - 81)
(x+9) / (x^2 - 81)
Factor the bottom...
(x+9) / [(x+9)(x-9)]
Cancel like terms...
1 / (x-9)
2007-12-13 14:17:50 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
the denominators are the same so u just need to work with the numerators...so the answer will be
(5x+2-4x+7)/(x^2-81)
(x+9)/(x+9)(x-9)
1/(x-9)
(x^2-81) is a difference of squares or if u don't know that just factor it out to get (x+9)(x-9)
hope this helps
2007-12-13 22:21:25 · answer #2 · answered by Megan 2 · 0⤊ 0⤋
x^2-81 is a common denominator thus,
((5x+2)-(4x-7))/(x^2-81)
Simplifying the numerator,
(x+9)/(x^2-81)
factoring the denominator,
(x+9)/[(x+9)(x-9)]
dividing numerator and denominator gives,
1/(x-9)
2007-12-13 22:49:31 · answer #3 · answered by Romy C 5 · 0⤊ 0⤋
I THINK x+5 over x^2-81 or do you need only one x in the solution?
2007-12-13 22:20:04 · answer #4 · answered by Anonymous · 0⤊ 0⤋
[(5x+2)/(x^2-81)] - [(4x-7)/(x^2-81)]
= [(5x+2) - (4x-7)] / (x^2-81)
= 5x+2-4x+7 / (x^2-81)
= x+9/(x^2-81)
= x+9/[(x+9)(x-9)]
= 1/(x-9)
2007-12-13 22:24:30 · answer #5 · answered by Anonymous · 0⤊ 0⤋
O_O' *feels dizzy*
now my head hurts.....
2007-12-13 22:17:59 · answer #6 · answered by Anonymous · 0⤊ 0⤋