# Thread: One more I need help linear systems

1. ## One more I need help linear systems

On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

Solve the system
{x+2y-2z=-1
{x+3y+z=10
{2x+6y+2z=20

solutions:
a. (8c-23, -3c+10, c) for any real number c.
b. (8c-22, -3c+10, c) for any real number c.
c. (8c-22, -3c+11, c) for any real number c.
d. (8c-23, -3c+11, c) for any real number c.

2. Originally Posted by mike1
On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

Solve the system
{x+2y-2z=-1
{x+3y+z=10
{2x+6y+2z=20
.
[ 1 2 -2 -1]
[ 1 3 1 10]
[ 2 6 2 20]
Divide the third row by 2,
[ 1 2 -2 -1]
[ 1 3 1 10]
[ 1 3 1 10]
Subtract 2nd from 3rd and subtract 1st from 2nd
[ 1 2 -2 -1]
[ 0 1 3 11]
[ 0 0 0 0]
Multiply 2nd by (-2) add to 1st,
[1 0 -8 -23]
[0 1 3 11]
Thus,
x-8z=-23
y+3z=11
Solve,
x=8z-23
y=-3z+11
And "z" can be anything thus z=c
Thus,
(x,y,z)=(8c-23,-3c+11,c)

3. Originally Posted by mike1
On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it?

Solve the system
{x+2y-2z=-1
{x+3y+z=10
{2x+6y+2z=20

solutions:
a. (8c-23, -3c+10, c) for any real number c.
b. (8c-22, -3c+10, c) for any real number c.
c. (8c-22, -3c+11, c) for any real number c.
d. (8c-23, -3c+11, c) for any real number c.
I am trained to look first for what is being asked for in a problem before I start thinking for the solution.
So I applied that in your question/problem above.
Being asked for is Solve the system, and possible answers are in terms of "c".

Where is "c" in the system?

------------------
The system has 3 unknowns, x,y,z, and 3 equations.
Good.

But the 2nd and 3rd equations are one and the same--only wearing different clothes--so in reality there are only 2 independent equations with 3 unknowns.
No good.

For a system to be "solvable", or with unique answers/solution, the number of independent equations must equal the number of unknowns.

4. Hello, Mike!

Solve the system : .$\displaystyle \begin{array}{ccc}x+2y-2z &= \;-1 \\x+3y+z & =\;10 \\2x+6y+2z& = \:20\end{array}$

$\displaystyle \begin{array}{cc}a.\;(8c-23, -3c+10, c) & b.\;(8c-22, -3c+10, c) \\ c.\;(8c-22, -3c+11, c) & d.\;(8c-23, -3c+11, c)\end{array}$ . for any real number $\displaystyle c$.

Since the second and third equations are equivalent,
. . we have the system: .$\displaystyle \begin{array}{cc}x + 2y - 2x & =\:-1 \\ x + 3y + z & =\:10\end{array}\begin{array}{cc}(1)\\(2)\end{arra y}$

Subtract (1) from (2): .$\displaystyle y + 3z\:=\:11\quad\Rightarrow\quad y \:=\:-3z + 11$

Substitute into (1): .$\displaystyle x + 2(-3z + 11) - 2x \:=\:-1\quad\Rightarrow\quad x \:=\:8x - 23$

We have $\displaystyle x,\,y,\,z$ expressed as functions of $\displaystyle z$: . $\displaystyle \begin{array}{ccc}x \:= & 8z - 23 \\ y \:= & -3z + 11 \\ z \:= & z\end{array}$

On the right, replace $\displaystyle z$ with a parameter $\displaystyle c$.

And we have: .$\displaystyle \begin{Bmatrix}x \:= & 8c - 23 \\ y \:= & -3z + 11 \\ z \:= & c\end{Bmatrix}$ . . . answer $\displaystyle d$

This is a parametric representation of all the solutions to the system.