Variable Missing?

Question

How do you solve a question like this:
Solve this linear system:
a-b-c=-1
a+c=2
a-2b=-7

I`m used to solving it when there`s 3 of the same variables in each equation...these are mixed up. How do I do this

Thanks a lot :)

Answer 1

When the variable is missing in the equation you can rewrite it with the missing variable included but with a coeffecient of zero.

a - b - c = -1
a +0b+c = 2
a -2b+0c = -7

Answer 2

Remember that whenever you have 3 equations, u can solve for 3 variables

a-b-c=-1 (1)
a+c=2 (2)
a-2b=-7 (3)

Adding (1) and (2)
2a-b=1
or 4a-2b = 2
a-2b=-7 (3)
Subtracting,
3a=9 =>a=3
=>c=-1 [from (2)] and b = 5 [from (3)]

Answer 3

from 1st equation, get c wherein: c = a-b+1
substitute this to 2nd equation: a + (a-b+1) = 2 = 2a-b+1 = 2
combine similar terms: 2a-b=1, then b = 2a-1
substitute b to 3rd equation: a-2(2a-1) = -7 = a - 4a + 2 = -7
combine similar terms: -3a = -9, therefore a = 3

since b = 2a-1, substitute value of a you get b = 2(3)-1
therefore b=5

since c = a-b+1 substitute both values of a and b you get
c = (3) - (5) + 1
c = -1

ans: a = 3, b = 5, c = -1

hope this helps

Answer 4

a-b-c=-1
c = 2 - a
b = (7 + a) / 2

then substitute the two lower equations into the first one and then solve to find out what a is. Then go back and figure out b and c knowing a.

a - ((7 + a) / 2) - (2 - a) = -1
a - (7 - a) / 2) - 2 + a = -1
2a - (7 - a) / 2) = 1
4a - 7 - a = 2
3a = 9
a = 3
therefore b = 5 and c = -1. put the values for a,b, and c back into the original equation to verify you got it right though.

Answer 5

call them a), b), c), and you do what you normally do!

a) a-b-c=-1
b) a +c=2, add
-------------
d) 2a -b =1
c) a-2b=-7 mult by 2
2a - 4b = -14, subtract d)
- 2a -b = 1
-------------
- 3b = -15
b = 5, sibs in c)

a - 2(5) = -7 or
a - 10 = -7 or
a = 3. subs in b)

3+c = 2 or
c = -1.

Check in original equation say a)
a -b -c =? -1
3 -5 - -1 =? -1
-2 + 1 = -1.

Done.

Answer 6

a-b-c=-1 <1
a+c=2 <2
a-2b=-7 <3
---------------------------------------
from eqn 2,
c=2-a
from eqn 3,
b=(7+a)/2
-------------------------------------------------
so,
a-[(7+a)/2]-(2-a)=-1
multiply throughout by 2,
[2a-7-a-4+2a]/2=-2
3a-11=-2 x 2
3a=-4+11
a=7/3
-------------------------
b=[7+(7/3)]/2
b=14/3
-------------------------
c=2-(7/3)
c=-1/3