2007-09-14 18:13:33 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
|10 + 3x| + 1 > 2
|10 + 3x| > 2 - 1
|10 + 3x| > 1
Two solutions:
(soln 1): 10 + 3x > 1
3x > 1 - 10
3x > -9
x > -3
(soln 2): -(10 + 3x) > 1
10 + 3x < -1
3x < -1 - 10
3x < -11
x < -11/3
2007-09-14 18:21:15 · answer #1 · answered by Pakyuol 7 · 2⤊ 0⤋
When you begin with a greater than sign, first remove the absolute value sign and solve regularly. Then, solve again, except the second time, flip the inequality sign to less than and change the sign of the number to the right of the inequality sign (if the number is positive it now becomes negative). Then compute.
Our two answers are: x > -3 OR x < -13/3
2007-09-14 18:24:10 · answer #2 · answered by sarasingergirl 3 · 0⤊ 0⤋
First add one to both sides. Then make two different equalititex. 10+3x >2 and 10 + 3x <-3 then solve for each equation.
2007-09-14 18:21:38 · answer #3 · answered by joe_89_9 4 · 0⤊ 0⤋
split the absolute function into the positive and negative solutions. so;
(10 + 3x) + 1 > 2
and
-(10 +3x) + 1 > 2 => -10 -3x + 1 > 2
I assume you can solve the rest. hope that helps. gtgkkbye.
2007-09-14 18:20:55 · answer #4 · answered by Anonymous · 0⤊ 0⤋
First get the equation in the absolute value brackets (|10 + 3x|) all by itself by subtracting one on both sides of the entire equation
Next, you will have either two solutions: 10 + 3x > 1 or 10 + 3x > -1
Now solve
And check your answers. Try the first answer. Does it fit the equation? If not then eliminate it and try the other one.
Done!
2007-09-14 18:24:33 · answer #5 · answered by Anonymous · 1⤊ 1⤋
10 + 3x +1 > 2
3x > 2 -10 -1
3x > -9
x > -3
2007-09-14 18:43:03 · answer #6 · answered by Freesia 5 · 0⤊ 1⤋
44
2007-09-17 09:33:51 · answer #7 · answered by simba w 1 · 0⤊ 1⤋
| 10 + 3x | > 1 gives:-
10 + 3x > 1
3x > - 9
x > - 3
AND
10 + 3x < - 1
3x < - 11
x < - 11 / 3
2007-09-18 11:09:38 · answer #8 · answered by Como 7 · 1⤊ 0⤋
i believe that's
x >-3
but i'm not sure..
C:
2007-09-14 18:22:56 · answer #9 · answered by Persy 3 · 1⤊ 0⤋