# Thread: Help: Solve the following systems, where a, b, and c are constants.

1. ## Help: Solve the following systems, where a, b, and c are constants.

Mods plz remove this one, i have posted in the right section now, i didn't want a advanced linear algebra answer.

This is a linear algebra question and i dont know how to solve for this because of the a,b, and c. i tried to do something but i dont know if its right.

I get the answer for part b as
x1 = a - c/3
x2 = -2b + b + c/3
x3 = 2a - b

Plz help me out and try to explain a bit if you can. I need help on a and b both.

2. Solve the following

$\displaystyle \left[ \begin{array}{c} x_1\\ x_2\\ x_3\end{array}\right]=\left[ \begin{array}{ccc} 1 & 1 & 1 \\ 2 & 0 & 2 \\ 0 & 3 & 3 \end{array} \right]^{-1} \left[ \begin{array}{c} a\\ b\\ c\end{array}\right]$

3. we are suppose to do it using the row-echlon form, i used reduced row echlon since it easier most of the time and we have only been taught till reduced row-echlon form. so i dont know what you mean by the -1 there because we didn't learn that yet.

4. ## Help: Solve the following systems, where a, b, and c are constants

This is a linear algebra question and i dont know how to solve for this because of the a,b, and c. i tried to do something but i dont know if its right.

I get the following answer using reduced row echlon for part b as
x1 = a - c/3
x2 = -2b + b + c/3
x3 = 2a - b

Plz help me out and try to explain a bit if you can. I need help on a and b both.

5. http://www.mathhelpforum.com/math-he...constants.html

You find the inverse of the matrix (the -1 part) by row echelon operations

6. Originally Posted by mastdesi
This is a linear algebra question and i dont know how to solve for this because of the a,b, and c. i tried to do something but i dont know if its right.

I get the following answer using reduced row echlon for part b as
x1 = a - c/3
x2 = -2b + b + c/3
x3 = 2a - b

Plz help me out and try to explain a bit if you can. I need help on a and b both.

(a) You can solve this by substitution, graphing or elimination.
Substitution:
$\displaystyle 2x + y = a \rightarrow y = -2x + a$
plug $\displaystyle y = -2x + a$ into $\displaystyle 3x + 6y = b$ so ...
$\displaystyle 3x + 6(-2x + a) = b$

Solve