# Intermediate algebra - "Solve each systems"

• Jan 23rd 2010, 10:26 PM
Justiceapple
Intermediate algebra - "Solve each systems"
Hi, I've been trying to solve this for over an hour now by extending the method of elimination like the textbook says. Very frustrated. (Headbang)

Solve each systems. Express the solution in the form (x,y,z,w).

x+y+z-w=5
2x+y-z+w=3
x-2y+3z+w=18
-x-y+z+2w=8

I don't know how to get there.
• Jan 23rd 2010, 11:42 PM
Prove It
Quote:

Originally Posted by Justiceapple
Hi, I've been trying to solve this for over an hour now by extending the method of elimination like the textbook says. Very frustrated. (Headbang)

Solve each systems. Express the solution in the form (x,y,z,w).

x+y+z-w=5
2x+y-z+w=3
x-2y+3z+w=18
-x-y+z+2w=8

I don't know how to get there.

$x + y + z - w = 5$
$2x + y - z + w = 3$
$x - 2y + 3z + w = 18$
$-x - y + z - 2w = 8$.

Apply $R_2 - 2R_1 \to R_2, R_3 - R_1 \to R_3, R_4 + R_1 \to R_1$.

$x + y + z - w = 5$
$-y - 3z + 3w = -7$
$-3y + 2z + 2w = 13$
$2z - 3w = 13$

Apply $-R_2 \to R_2$

$x + y + z - w = 5$
$y + 3z - 3w = 7$
$-3y + 2z + 2w = 13$
$2z - 3w = 13$

Apply $R_3 + 3R_2 \to R_3$

$x + y + z - w = 5$
$y + 3z - 3w = 7$
$11z - 7w = 34$
$2z - 3w = 13$

Interchange $R_3$ and $R_4$.

$x + y + z - w = 5$
$y + 3z - 3w = 7$
$2z - 3w = 13$
$11z - 3w = 13$

Apply $\frac{1}{2}R_3$

$x + y + z - w = 5$
$y + 3z - 3w = 7$
$z - \frac{3}{2}w = \frac{13}{2}$
$11z - 3w = 13$

Apply $R_4 - 11R_3$

$x + y + z - w = 5$
$y + 3z - 3w = 7$
$z - \frac{3}{2}w = \frac{13}{2}$
$\frac{27}{2}w = -\frac{117}{2}$

So $w = \frac{2}{27}\left(-\frac{117}{2}\right)$

$w = -\frac{13}{3}$.

$z - \frac{3}{2}w = \frac{13}{2}$

$z - \frac{3}{2}\left(-\frac{13}{3}\right) = \frac{13}{2}$

$z + \frac{13}{2} = \frac{13}{2}$

$z = 0$.

$y + 3z - 3w = 7$

$y + 3(0) - 3\left(-\frac{13}{3}\right) = 7$

$y + 13 = 7$

$y = -6$.

$x + y + z - w = 5$

$x - 6 + 0 + \frac{13}{3} = 5$

$x - \frac{5}{3} = 5$

$x = \frac{20}{3}$.

To check, sub into original Eq 2:

$2x+y-z+w = 2\left(\frac{20}{3}\right) - 6 - 0 - \frac{13}{3}$

$= \frac{40}{3} - \frac{18}{3} - \frac{13}{3}$

$= \frac{9}{3}$

$= 3$ as required.

So $(x, y, z, w) = \left(\frac{20}{3}, 6, 0, -\frac{13}{3}\right)$.

The solution your text book gives is incorrect.
• Jan 24th 2010, 12:08 AM
Justiceapple
Oh no, you wrote down the problem wrong :(

The R4 is $-x-y+z+2w=8$. It's +2w.

But I think I kind of get what to do now. Let me work on this problem first thing in the morning tomorrow. My head hurts from studying Logic for another class right now....

Thank you so much. When I saw that you replied, I screamed like I won a lottery or something. I appreciate your kindness very much.

P.S. I studied Math for 6 hours today ...
• Jan 24th 2010, 12:11 AM
Prove It
Oh hahahaha, yes I copied it down wrong.

But you see the basic structure now.

Try to solve yours the same way.
• Jan 24th 2010, 12:14 AM
VonNemo19
Quote:

Originally Posted by Justiceapple
Oh no, you wrote down the problem wrong :(

(Rofl) That's pretty funny.
• Jan 24th 2010, 12:36 AM
Prove It
Quote:

Originally Posted by VonNemo19
(Rofl) That's pretty funny.

So I think the score is Prove It -1, Von Nemo -2 (Rock)