# How tdo I solve a Linear Equation with an unknown

1. I am having a trouble with this math problem and would like help finding the answer. Here goes: Let (the ~~ indicates where a line should straight line should be going downward and /// are like a space between the numbers)
1 /// (-1) /// 1 ~~ 2
a /// 0 /// a+1 ~~ 2a+3
2a /// (-2a) /// (a^2)-3 ~~ 5a-3

be the augmented matrix of a linear system. For what values of a does the system have no solution? One solution? Infinitely many solutions?

So, if I correctly reduced it, that would mean that
1 /// 0 /// 1+1/a ~~ 2+3/a
0 /// 1 /// 1/a ~~ 3/a
0 /// 0 ///(a-3)(a+1) ~~ a-3

0 solutions when a=-1
infinite solutions when a=3
one solution when a/=/(doest not equal) 1, 3
And also a/=/ 0 because that would be dividing by 0.

Is this the way you are supposed to do this type of problem?

I don't really understand the matrix. Can you not write it a little clearer? then maybe I can help.

Hmm... I don't really know how to write it clearer does anyone know how to write a matrix on a forum?

[tex ] \left(

\begin{array}{ccc|c}

a & b & c & u\\

d & e & f & v\\

g & h & j & w\\

\end{array}

\right)

[/ tex]

(except leave out the spaces in the "tex" tags. Substitute the values you need.

It looks like this:

$$\left( \begin{array}{ccc|c} a & b & c & u\\ d & e & f & v\\ g & h & j & w\\ \end{array} \right)$$

--Elucidus

The OP meant:
$$\begin{vmatrix} 1 & -1 & 1 & 2\\ a & 0 & a+1 & 2a+3 \\ 2a & -2a & a^2-3 & 5a-3 \end{vmatrix}$$

Anyway you did it right, except you should mention that for a=0 there are also 0 solutions.

Also, there is one solution when a/=/(doest not equal) -1, 3

I guess it was typo.