# Thread: three-variable solution using elimintation: I'm stuck!

1. ## three-variable solution using elimintation: I'm stuck!

I am helping my daughter and this one has me stumped - it is a three-variable equation and their asking her to use "elimination" to solve the problem. The text also states that "If you obtain a false equation, such as 0=1, in any of the steps, then the system has no solution. If you do not obtain a false equation, but obtain an identity such as 0=0, then the system has infinitely many solutions."

#13 in Mcougal Little Algebra 2, pp 182:

Eq#1: 5x + y - z = 6
Eq#2: x + y + z = 2
Eq#3: 3x + y = 4

My steps:
"bounce" EQ#1 of EQ#2 and eliminate the "z" variable:

5x + y - z = 6
x + y + z = 2
____________
6x + 2y = 8 (New EQ#1')

Bounce EQ#1' against EQ#3 and eliminate the x variable:

6x + 2y = 8 (EQ#1')
-2*(3x + y = 4)

yields...

6x + 2 y = 8
-6x - 2y = -8
_____________
0 + 0 = 0 (or 0=0, infinitely many solutions).

Did it several times, comes out the same.

Yet the answer in the back of the book gives (0, 4, -2) as the ordered triplet - I plug this into all three equations, and it works!

What gives? and where am I going wrong?

Thanks.

2. I have done some eliminations also and they all led to similar equations, hence, infinitely many solutions.

The book's (0,4,-2) is one of those many solutions. It is a "trick". It assigned a value to one of the 3 variables to get the other two variables. Maybe it let x = 0, to get the y and z
.... or let y = 4, to get the x and z
.... or let z = -2, to get the x and y

3. Originally Posted by taekwondodo
I am helping my daughter and this one has me stumped - it is a three-variable equation and their asking her to use "elimination" to solve the problem. The text also states that "If you obtain a false equation, such as 0=1, in any of the steps, then the system has no solution. If you do not obtain a false equation, but obtain an identity such as 0=0, then the system has infinitely many solutions."

#13 in Mcougal Little Algebra 2, pp 182:

Eq#1: 5x + y - z = 6
Eq#2: x + y + z = 2
Eq#3: 3x + y = 4

My steps:
"bounce" EQ#1 of EQ#2 and eliminate the "z" variable:

[quote]5x + y - z = 6
x + y + z = 2
____________
6x + 2y = 8 (New EQ#1')[\quote]
Yes, that is correct. And, as you note below, 6x+ 2y= 8 is really the same equation (since you can multiply or divide an equation by any non-zero number and get an "equivalent" equation) as 2(3x+ y)= 2(4) or 3x+ y= 4, your third equation.

That is, you really have only two INDEPENDENT equations and so cannot solve for a specific value of (x,y,z).

Geometrically, you can think of each equation as representing a plane in an xyz- coordinate system. Two planes may intersect in an entire line, not just in one point. From 3x+ y= 4, y= 4- 3x. Then x+ y+ z= 2 becomes x+ (4- 3x)+ z= 2 or z= 2-x-(4-3x)= 2- x- 4+ 3x= 2x- 2. Any point on the line given by (x, 4- 3x, 2x-2) will lie on the intersection of both planes.

Bounce EQ#1' against EQ#3 and eliminate the x variable:

6x + 2y = 8 (EQ#1')
-2*(3x + y = 4)

yields...

6x + 2 y = 8
-6x - 2y = -8
_____________
0 + 0 = 0 (or 0=0, infinitely many solutions).

Did it several times, comes out the same.

Yet the answer in the back of the book gives (0, 4, -2) as the ordered triplet - I plug this into all three equations, and it works!

What gives? and where am I going wrong?

Thanks.