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Help with #6

5CC99664-7D2B-49D4-96FE-020BA6F52321.jpg.jpeg
Reply 1
Avatar for Krab21
Krab21
8
a=1
b=3
c=0
Reply 2
Avatar for Ndanji678
Ndanji678
OP
9
Original post by Krab21
a=1
b=3
c=0

But how to get those answers?
Original post by Ndanji678
5CC99664-7D2B-49D4-96FE-020BA6F52321.jpg.jpeg


First write out the equations for the values of U1, U2 and U3:
1. 4 = a + b + c
2. 10 = 4a + 2b + c
3. 18 = 9a + 3b + c

Taking away #1 from the others gets you the following:
4. 6 = 3a + b
5. 14 = 8a + 2b --> 7 = 4a + b
So a = 1

Now you can work out the value of b. Using #4:
b + 3 = 6
So b = 3

Then using #1:
1 + 3 + c = 4
Therefore c = 0
Reply 4
Avatar for Ndanji678
Ndanji678
OP
9
Original post by SkyRunner61
First write out the equations for the values of U1, U2 and U3:
1. 4 = a + b + c
2. 10 = 4a + 2b + c
3. 18 = 9a + 3b + c

Taking away #1 from the others gets you the following:
4. 6 = 3a + b
5. 14 = 8a + 2b --> 7 = 4a + b
So a = 1

Now you can work out the value of b. Using #4:
b + 3 = 6
So b = 3

Then using #1:
1 + 3 + c = 4
Therefore c = 0


Original post by SkyRunner61
First write out the equations for the values of U1, U2 and U3:
1. 4 = a + b + c
2. 10 = 4a + 2b + c
3. 18 = 9a + 3b + c

Taking away #1 from the others gets you the following:
4. 6 = 3a + b
5. 14 = 8a + 2b --> 7 = 4a + b
So a = 1

Now you can work out the value of b. Using #4:
b + 3 = 6
So b = 3

Then using #1:
1 + 3 + c = 4
Therefore c = 0


For the part where you’re saying you take away #1 from the others and get that, when you start subtracting don’t you get 8=5a+b?
Original post by Ndanji678
For the part where you’re saying you take away #1 from the others and get that, when you start subtracting don’t you get 8=5a+b?


I mean...that is true but I chose to eliminate c by taking away #1 from the other two individually since you get to avoid dipping into negative numbers and it's the smallest value to subtract
Reply 6
Avatar for Ndanji678
Ndanji678
OP
9
Original post by Ndanji678
For the part where you’re saying you take away #1 from the others and get that, when you start subtracting don’t you get 8=5a+b?


Original post by SkyRunner61
First write out the equations for the values of U1, U2 and U3:
1. 4 = a + b + c
2. 10 = 4a + 2b + c
3. 18 = 9a + 3b + c

Taking away #1 from the others gets you the following:
4. 6 = 3a + b
5. 14 = 8a + 2b --> 7 = 4a + b
So a = 1

Now you can work out the value of b. Using #4:
b + 3 = 6
So b = 3

Then using #1:
1 + 3 + c = 4
Therefore c = 0


I figured it out,thanks.

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