2006-12-06 13:06:01 · 4 answers · asked by 2thpick 1 in Science & Mathematics ➔ Mathematics
-12 < x < 6
-9 < x+3< 9
(^ that's the trick)
|x+3| < 9
another way to look at it. You want the midpoint to be the symmetric point (-12+6)/2 = -3.
2006-12-06 13:12:38 · answer #1 · answered by modulo_function 7 · 1⤊ 0⤋
What you have to do is get the value smack in the middle of them. Take the average of the two numbers.
(-12 + 6 )/2 = -3
Now, determine the distance -3 is away from 6. The answer is 9.
Take the equation |x - c| = d, where d = the distance (which is 9), and c = the average (-3)
Therefore |x + 3| = 9
This implies x + 3 = 9 and x + 3 = -9 are solutions, or x = 6 and x = -12
2006-12-06 13:14:24 · answer #2 · answered by Puggy 7 · 0⤊ 0⤋
The pair might have distinctive suggestions if the two equations at the instant are not linearly self sufficient. For that to be the case, one equation is comparable to the different prolonged via a persevering with: (ok-3)x + 3y - 6 = akx + aky - 12a = 0 equate coefficients ok-3 = ak 3 = ak 6 = 12a from the final, a = 0.5; from the previous, ok = 6; examine with the 1st ok-3 = 3 and ak = 3. So the situation for distinctive solutions is okay = 6.
2016-12-11 03:45:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
- l12l = -12
l6l = 6
2006-12-06 13:27:43 · answer #4 · answered by Anonymous · 0⤊ 1⤋