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I have spent all morning trying to solve this. :banghead:

Problem #1 from section 4-6 (Systems of Linear Equations with Three or More Variables)

x-2y+3z=3

2x+y+5z=8

3x-y-3z=-22

Neither ds nor I can figure it out and neither can our math friend who lives down the street. Please help!

Thanks.

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I have a solutions manual, if you want me to find it for you....Char

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x-2y+3z=3

2x+y+5z=8

3x-y-3z=-22

Okay, here goes: Start by getting two equations where the coefficient of y = 0.

Multiply the second equation by 2, then add the resulting equation to the first equation.

2(2x+y+5z) = 16

4x+2y+10z = 16 (add it to first equation)

x-2y+3z=3 getting

5x+13z=19

Now get a similar equation by adding the second and third equations:

5x+8z=-14

Using those two equations without y, solve for x and z:

5x+13z=19

5x+2z=-14 (subtract second from first)

11z=33; z= 3; so x = -4

Then plug those into one of the equations to solve for y. Let's use the first one: x-2y+3z=3.

So -4-2y+9=3

-2y+5 = 3

-2y=-2; y=1

Solution set is {(-4,1,3)}

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You have three linear equations in three unknowns: the standard procedure is to start by eliminating one of the variables. Let's choose "x" to eliminate:

Multiply Eqn #1 by -3 and add the result to Eqn #3:

-3x + 6y - 9z = -9

3x - y - 3z = -22

0 + 5y - 12z = -31

Now multiply Eqn #1 by -2 and add the result to Eqn #2:

-2x + 4y -6z = -6

2x + y +5z = 8_

0 + 5y - z = 2

Now we're down to two equations in two unknowns (y & z):

5y -12 z =-31

5y - z = 2

Eliminate y by multiplying the first equation by -1 and adding to the second equation. This allows you to solve for z:

-5y + 12z = 31

+5y - z = 2

0 + 11 z = 33

Solving for z:

z = 33/11

z = 3

Now solve for y by going back to 5y - z = 2 and plugging in z=3:

5y - 3 = 2

5y = 5

y = 1

Finally, use y=1 and z=3 in one of the original equations and solve for x:

x - 2y + 3z = 3

x - 2*1 + 3*3 = 3

x - 2 + 9 = 3

x = +2 - 9 + 3

x = -4

Done!

ETA: This method is the same as Gailmegan's method, only she eliminated "y" first and was left with two equations in x and z to solve.

Edited by Kathy in Richmond
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The main idea to get across to the students is that they must reduce the system to equations with the same two variables. Novices sometimes eliminate one variable the first time and a different variable the second time....

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The main idea to get across to the students is that they must reduce the system to equations with the same two variables. Novices sometimes eliminate one variable the first time and a different variable the second time....

:iagree:

Also, while you can use any two equations for the first elimination, the second set MUST use the third equation that wasn't used...and one that's been used.

So in Kathy's example, she first used eqn 1 and 3,

Her next elimination MUST use eqn 2 (otherwise you end up with the same equation). So use eqn 2 and either eqn 1 or 3...whichever works better.

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You have three linear equations in three unknowns: the standard procedure is to start by eliminating one of the variables. Let's choose "x" to eliminate:...

ETA: This method is the same as Gailmegan's method, only she eliminated "y" first and was left with two equations in x and z to solve.

We must have been solving at the same time. Yours looks prettier though; I couldn't figure out how to format it to make it look right. :D

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Thank you!!!!!! I was making it so much more complicated than it needed to be. After I reduced the equations to two variables, I was starting all over trying to find the other variable rather than just plugging it in <blush>. I hope that Algebra 2 will not be the death of me this year!

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