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Papessa [141]
3 years ago
9

What is the solution to the following system ?

Mathematics
1 answer:
zvonat [6]3 years ago
4 0

3x+3y+6z=9
x+3y+2z=5
3x+12y+12z=18


Step 1: use Eq 2 to solve for x
x+3y+2z=5
x= -3y-2z+5
Step 2: Solve for y unsing Eq 1
3x+3y+6z=9
3y= -3x-6z+9
y= -x-2z+3

Step 3: plug in x= -3y-2z+5 to solve for the value of y
y=-x-2z+3
y= (-1)(-3y -2z +5) - 2z+3
y= 3y+2z-5- 2z+3
-2y= -2
y=1

Step 4: plug y= 1 into the x= equation 
x= -3y-2z+5
x= -3(1)-2z+5
x= -2z+2
Step 5:Plug y= 1 and x= -2z+2 into the 3rd equation to solve for z
3x+12y+12z=18
12z= -3(-2z+2)-12(1)+ 18
12z= 6z - 6 - 12 +18
6z= -18+18
6z=0
z=0

Step 6: Plug z in to solve for x
x=-2z+2
x= (-2)(0)+2
x= 2

ANSWER:
x,y,z = 2,1,0
answer choice C

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Find an explicit formula for the arithmetic sequence -11, -3, 5, 13, ....
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Answer:

<u>b(n) = -11 + (n-1)*8</u>

<u></u>

Step-by-step explanation:

Let n be the sequence number, with n=1 the first number b(1) -11

The sequence changes by +8 each step.

<u><em>b(1)</em></u><em>   -11</em>         + 8 =   -3

<u><em>b(2)</em></u><em>   -3</em>          + 8 =    5

<u><em>b(3)</em></u>   <em> 5 </em>         + 8 =   13

<u><em>b(4)</em></u><em> </em> <em>13</em>

b(1) = -11

b(2) = -3.   or b(1) + 1*8

b(3) =  5,    or b(1) + 2*8

b(4) =  13,   or b(1) + 3*8

We note that each step adds a multiple of 8 to the initial value of -11.  This can be stated as (n-1)*8

The formula for this sequence would be b(n) = b(1) + (n-1)*8

<u>b(n) = -11 + (n-1)*8</u>

Check:  Does n=3 return the value of 5?

b(n) = -11 + (n-1)*8

b(3) = -11 + ((3)-1)*8

b(3) = -11 + (2)*8

b(3) = -11 + 16

b(3) = 5   <u><em>YES</em></u>

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