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

Printable View

- September 23rd 2006, 06:29 PMimthatgirllinear 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 - September 23rd 2006, 06:41 PMQuick
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 - September 24th 2006, 01:00 AMclassicstrings
- September 24th 2006, 01:30 AMSoroban
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)

- September 24th 2006, 05:07 PMThePerfectHacker
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.