i have an algebra final, and i'm blanking out on everything. first person to show me how to do this problem gets best answer. i promisee <3 so please help me out.
x
------
x + 1
+
3
------
x - 3
+
1
=
0
^ sorry i had to write it vertically, because it didn't show up right horizontally!
2007-05-28 15:11:31 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
your common denominator is (x+1)(x-3), so
multiply the first by (x-3)
x(x-3)+3(x+1)+(x+1)(x-3)=0
x^2-3x+3x+3+x^2-3x+x-3=0
2x^2-2x=0
2x(x-1)=0
x=0 x=1
2007-05-28 15:16:08 · answer #1 · answered by leo 6 · 1⤊ 1⤋
can not do a million, yet 2-30: 2. C 3. D 4. A 5. A 6. D 7. C 8. A 9. D 10. B 11. A 12. A 13. A 14. a fifteen. C sixteen. C 17. D 18. D 19. A 20. A 21. D 22. B 23. A 24. B 25. A 26. A 27. B 28. A 29. B 30. B
2016-11-23 12:48:58 · answer #2 · answered by hundson 4 · 0⤊ 0⤋
x/(x+1) + 3/(x-3) + 1 = 0
First multiply by (x+1)(x-3) to get rid of all the denominators:
x(x-3) + 3(x+1) + (x+1)(x-3) = 0
x^2 - 3x + 3x + 3 + x^2 - 2x - 3 = 0
2x^2 -2x = 0
x (x-1) = 0
x = 0 or x = 1
2007-05-28 15:16:35 · answer #3 · answered by McFate 7 · 1⤊ 2⤋
x/(x+1) + 3/(x-3) + 1 = 0
LCD: (x+1)(x-3)
[x(x-3) + 3(x+1) + (x+1)(x-3)]/[(x+1)(x-3)] = 0
Numerator must equal 0
x^2 - 3x + 3x + 3 + x^2 - 2x - 3 = 0
2x^2 - 2x = 0
x^2 - x = 0
x(x-1) = 0
x = 0
x = 1
2007-05-28 15:18:55 · answer #4 · answered by gudspeling 7 · 0⤊ 0⤋
hey u need not worry..
consider the first 2.
take L.C.M
(x/x+1 )+(3/x-3) =[(x*(x-3))+(3*(x+1))]/(x+1)(x-3)
then it wud be (x2-3x+3x+3)/(x+1)(x-3)
ie (x2+3)/x2+x-3x-3=(x2+3)/x2-2x-3
so add 1 now.
it wud be x2+3+x2-2x-3=o
(becoz if u bring the denominator to the right it becomes 0)
so ur equation is 2x2-2x=0
2(x2-x)=0
x2-x=0; x(x-1)=0;
so x=1 or x=0
2007-05-28 15:22:58 · answer #5 · answered by physics 2 · 0⤊ 0⤋
The solution set is x=(0,1)
2007-05-28 15:21:45 · answer #6 · answered by Sciencenut 7 · 0⤊ 2⤋