Asked by Saphina AlMatary
Solve the simultaneous questions:
(a) x+y+z=-3 (1)
x-y-2z=13 (2)
y-4z=25 (3)
(b)x+2y+3z=-14 (1)
2x+3y-z=13 (2)
3x-y+2z=-7 (3)
(a) x+y+z=-3 (1)
x-y-2z=13 (2)
y-4z=25 (3)
(b)x+2y+3z=-14 (1)
2x+3y-z=13 (2)
3x-y+2z=-7 (3)
Answers
Answered by
Damon
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
for the first one for example there are three rows and four columns
1 +1 +1 -3
1 -1 -2 13
1 +0 -4 25
for the first one for example there are three rows and four columns
1 +1 +1 -3
1 -1 -2 13
1 +0 -4 25
Answered by
Damon
x = 15/7
y = 4/7
z = -40/7
y = 4/7
z = -40/7
Answered by
Reiny
There is no unique or best way to solve these.
After doing a few you get a "feel" of what to look for.
e.g. for your first one, notice that last one says:
y - 4z = 25 , so y = 25+4z
- could be useful
look at the first two, they both have 1x
so lets subtract the 2nd from the 1st:
2y + 3z = - 16
remember y = 25+4z
2(25+4z) + 3z = -16
50 + 8z + 3z = -16
11z = -66
z = -6
then y = 25 + 4(-6) = 1
back in the 1st:
x+y+z=-3
x+1-6=-3
x = 2
x=2 , y=1 , z=-6
hint for the 2nd:
double the first and subtract from the 2nd to get a
y , z equation
triple the first and subtract from the 3rd to get another y , z equation
now solve as 2 equations in two unknowns
After doing a few you get a "feel" of what to look for.
e.g. for your first one, notice that last one says:
y - 4z = 25 , so y = 25+4z
- could be useful
look at the first two, they both have 1x
so lets subtract the 2nd from the 1st:
2y + 3z = - 16
remember y = 25+4z
2(25+4z) + 3z = -16
50 + 8z + 3z = -16
11z = -66
z = -6
then y = 25 + 4(-6) = 1
back in the 1st:
x+y+z=-3
x+1-6=-3
x = 2
x=2 , y=1 , z=-6
hint for the 2nd:
double the first and subtract from the 2nd to get a
y , z equation
triple the first and subtract from the 3rd to get another y , z equation
now solve as 2 equations in two unknowns
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