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
Hello, imthatgirl!
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)
Not always.
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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.