# linear equations

• Sep 23rd 2006, 06:29 PM
imthatgirl
linear equations
can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
2a-5c=-18
7b-8c=-65
3a-2b=2
• Sep 23rd 2006, 06:41 PM
Quick
Quote:

Originally Posted by imthatgirl
can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
2a-5c=-18
7b-8c=-65
3a-2b=2

you have: 2a-5c=-18

then: 2a=5c-18

thus: a=(5/2)c-9

You also have equation: 3a-2b=2

substitute: 3((5/2)c-9)-2b=2

then: (15/2)c-27-2b=2

then: -2b=2+27-(15/2)c

then: b=(15/4)c-(27/2)-1

You also have equation: 7b-8c=-65

substitute: 7((15/4)c-(27/2)-1)-8c=-65

then: (105/4)c-(189/2)-8c=-65

now you have an equation with one variable...
make sure to check my arithmetic
• Sep 24th 2006, 01:00 AM
classicstrings
Quote:

Originally Posted by imthatgirl
can anyone explain how to slove this? there's 3 equations and 3 variables, but only 2 variables in each equation
2a-5c=-18
7b-8c=-65
3a-2b=2

3 equations, 3 unknowns = solvable.
3 equations, 4 unknowns = a system of equations* (check) i.e. many solutions

4 equations, 4 unknowns = solvable
... and so on
• Sep 24th 2006, 01:30 AM
Soroban
Hello, imthatgirl!

Quote:

Can anyone explain how to solve this?
There's 3 equations and 3 variables, but only 2 variables in each equation.

. . 2a - 5c .= .-18
. . 7b - 8c .= .-65
. . 3a - 2b .= . 2

Missing variables should not be a problem.
If you're using Elimination, they actually did us a favor!
Code:

(1)  2a      - 5c  =  -18
(2)        7b - 8c  =  -65
(3)  3a - 2b      =    2

Multiply (1) by -3, multiply (3) by 2, and add:
Code:

-3(1)      -6a      + 15c  =  54
2(3)      6a - 4b        =  4
- - - - - - - - - - -
(4)      - 4b + 15c  =  58

Multiply (4) by 7, multiply (2) by 4, and add:
Code:

7(4)    -28b + 105c  =  406
4(2)      28b -  32c  = -260
- - - - - - - - - -
73c  =  146

And we have: .c = 2

Substitute into (1): . 2a - 5·2 .= .-18 . . a = -4

Substitute into (2): . 7b - 8·2 .= .-65 . . b = -7

Therefore: .(a,b,c) .= .(-4,-7,2)

• Sep 24th 2006, 05:07 PM
ThePerfectHacker
Quote:

Originally Posted by classicstrings
3 equations, 3 unknowns = solvable.
3 equations, 4 unknowns = a system of equations* (check) i.e. many solutions

4 equations, 4 unknowns = solvable
... and so on

Not always.
---
Firstly, we for simplicity sake are speaking about linear equations. [Otherwise, the mathematics that studies ploynomial systems is called algebraic geometry, we are speaking about linear algebra-meaning linear systems].

The necessary and sufficient condition for a n variable n equations system have a unique solution is that the determinant is non-zero.

For example,
x+y=1
x+y=2

Has the solution set as an empty set.