I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
Answer:
1/2
((-2)^2-(4*2)^1/3)/abs(-2*2)
-2^2 = 4
4*2 = 8
-2*2 = -4 and abs of that is 4
4-(8)^1/3/4
8^1/3 = 2
4-2 = 2
2/4 = 1/2
Step-by-step explanation:
Answer:
#3×|x-9|=0
|x-9|=0
X-9=0
X=9
# |2x-9|-4=2
|2x-9|=2+4
|2x-9|=6
2x-9=6
2x=9=-6
X=15/2
X=3/2
1 to x power =3/2 or
2 to x power =15/2
Step-by-step explanation:
Hope I helped u
