Solving a system of linear equations

• Specter
In summary: I would have done it that way, too, but the approach I suggested is more general, and you need not even substitute out the ##x## term, as you can proceed directly to the final equation, if you like. It is not a matter of computing more or less, it is a matter of having to think less. It is also a matter of being able to easily automate the solution process. This is important for large numbers of variables, as well.
Specter

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

n1⋅(n2×n3)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n1=(1,1,-9)
n2=(1,2,3)
n3=(2,-3,0)

n1⋅(n2×n3)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

Specter said:

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

n1⋅(n2×n3)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n1=(1,1,-9)
n2=(1,2,3)
n3=(2,-3,0)

n1⋅(n2×n3)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

12z − ( −9z) 3z

Specter said:

Homework Statement

Determine whether the following system of equations has a single point of intersection. If so, find the point of intersection.

4x+y-9z=0
x+2y+3z=0
2x-3y-5=0

n1⋅(n2×n3)

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.

n1=(1,1,-9)
n2=(1,2,3)
n3=(2,-3,0)

n1⋅(n2×n3)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.4x+y-9z=0 [1]
x+2y+3z=0 [2]
2x-3y-5=0 [3]

4x+y-9z=0 [1]
4x+8y+12z=0 [4] Eqn [2] x 4 to eliminate x.
Subtract and the new equation is 7y+3z=0 [5].

Use a different pair of equations to eliminate x. I used equations 1 and 3.

4x+y-9z=0 [1]
4x-6z-10=0 [6] (Eqn 3x2)
Subtract and the new equation is -5y-9z-10=0 [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.

Use Gaussian elimination; that is the standard method we almost always apply in such problems. It does not matter if you have 2 variables, 3 variables or 100 variables---the method does not change; only the amount of work is different in such cases.

Anyway, if your third equation really is ##2x-3y=5-0## (exactly as written), you get one solution, but if it was supposed to be ##2x-3y-5z=0## you get a different solution. Both problems can be tackled successfully via Gaussian elimination.

Last edited:
You can get the relation ##x = \frac{5+3y}{2}##,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Last edited:
Young physicist said:
Another way to solve this is by substitution:
You can get the relation ##x = \frac{5+3y}{2}##,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Yes, indeed, and that is what is meant by "Gaussian elimination".

Ray Vickson said:
Yes, indeed, and that is what is meant by "Gaussian elimination".
Oops! My previous post edited.

Young physicist said:
You can get the relation ##x = \frac{5+3y}{2}##,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.
Thanks! I'll try it.

Ray Vickson said:
Use Gaussian elimination; that is the standard method we almost always apply in such problems. It does not matter if you have 2 variables, 3 variables or 100 variables---the method does not change; only the amount of work is different in such cases.

Anyway, if your third equation really is ##2x-3y=5-0## (exactly as written), you get one solution, but if it was supposed to be ##2x-3y-5z=0## you get a different solution. Both problems can be tackled successfully via Gaussian elimination.

Thank you. And yes the third equation is 2x-3y=5-0. I'll try to do it the way you have told me but I haven't learned it like that so I might run into some trouble.

Young physicist said:
You can get the relation ##x = \frac{5+3y}{2}##,
from the last equation ,then you can substitute x away in the two previous equations and get two equations each with two variables.

Then you can do the process again,and get the final equation with only one variable.then use the relations to find the other variables.

Whats the reason in choosing eqn [3] and not [1] or [2] to solve for x? How do I know which of the equations to choose when solving the question using this method?

Specter said:
Whats the reason in choosing eqn [3] and not [1] or [2] to solve for x? How do I know which of the equations to choose when solving the question using this method?
Eqn 3 is the only equation in the system that has exact two variables,so it’s the only equation that can give the relation between two and only two variables.

Specter
Young physicist said:
Eqn 3 is the only equation in the system that has exact two variables,so it’s the only equation that can give the relation between two and only two variables.

Thanks! So I solved for x the way I was told to, getting x=5+3y/2

Next I substitute the x value into equation [2] so solve for z, giving me z=- 5+7y/6

Then I substitute the x and z values into equation [1] to solve for y, giving me y=-1.

I'm not sure what to do next, I see people getting different answers than me. Is there more to the question?

Edit: Never mind I just have to substitute the values into x and z to get my coordinates.

Specter said:
Thanks! So I solved for x the way I was told to, getting x=5+3y/2
This is incorrect.
From your third equation, 2x - 3y = 5, solving for x gives 2x = 3y + 5, so ##x = \frac{3y + 5} 2##. Maybe that's what you meant, but what you wrote above says ##x = 5 + \frac{3y} 2##, and this is different from what I wrote.

Specter said:
Next I substitute the x value into equation [2] so solve for z, giving me z=- 5+7y/6

Then I substitute the x and z values into equation [1] to solve for y, giving me y=-1.

I'm not sure what to do next, I see people getting different answers than me. Is there more to the question?
I get a different value for y, not -1.
Edit: Your value for y is correct. I missed a minus sign when I wrote down the original system of equations.
Once you get a solution, you can (and should) substitute the three values in your original system of equations, from which you should get three true statements.

Last edited:
YoungPhysicist and Specter
Mark44 said:
This is incorrect.
From your third equation, 2x - 3y = 5, solving for x gives 2x = 3y + 5, so ##x = \frac{3y + 5} 2##. Maybe that's what you meant, but what you wrote above says ##x = 5 + \frac{3y} 2##, and this is different from what I wrote.I get a different value for y, not -1.

Once you get a solution, you can (and should) substitute the three values in your original system of equations, from which you should get three true statements.
I keep getting y=-1, using ##x = \frac{5 + 3y} 2## and ##z= - \frac{5+7y} 6##

I can show my work if that would help! I haven't used LaTeX before... is there an extension for chrome or is it used by typing out the code?

Thanks

Specter said:
I keep getting y=-1, using ##x = \frac{5 + 3y} 2## and ##z= - \frac{5+7y} 6##
Your value for x is correct. I missed a sign when I wrote down your original system.

What values do you get for x and z?
Specter said:
I can show my work if that would help! I haven't used LaTeX before... is there an extension for chrome or is it used by typing out the code?
There's no extension -- you write down the LaTeX script. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/

Specter
Mark44 said:
Your value for x is correct. I missed a sign when I wrote down your original system.

What values do you get for x and z?
There's no extension -- you write down the LaTeX script. See our tutorial here -- https://www.physicsforums.com/help/latexhelp/

Thanks. I'll take a look.

I got x=1 and ##z= \frac{1} 3##

So the point of intersection would be ##(1,-1,\frac{1} 3##)

Specter said:
Thanks. I'll take a look.

I got x=1 and ##z= \frac{1} 3##

So the point of intersection would be ##(1,-1,\frac{1} 3##)
Yes, that's the solution, which you can check.
Is 4(1) + 1(-1) -9(1/3) = 0?
Is 1 + 2(-1) + 3(1/3) = 0?
Is 2(1) - 3(-1) = 5?
If all three are true, you've found the solution.

Specter
Specter said:
Determine whether the following system of equations has a single point of intersection.
Did you do this part? Possibly it entails finding the determinant of the matrix of coefficients, but I don't know if you have covered that yet.

Mark44 said:
Did you do this part? Possibly it entails finding the determinant of the matrix of coefficients, but I don't know if you have covered that yet.
Oh I'm pretty sure what my teacher means by determine if there is a single point of intersection was this:

Specter said:
n1=(1,1,-9)
n2=(1,2,3)
n3=(2,-3,0)

n1⋅(n2×n3)

=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.

Thats all I can think of because I haven't learned what you just mentioned.

Specter said:
Oh I'm pretty sure what my teacher means by determine if there is a single point of intersection was this:

Specter said:
n1=(1,1,−9)
n2=(1,2,3)
n3=(2,−3,0)

n1⋅(n2×n3)

=(4,1,−9)⋅[(1,2,3)×(2,−3,0)]
=(4,1,−9)⋅(9,6,−7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.

That's all I can think of because I haven't learned what you just mentioned.
The product you calculated, known as the scalar triple product, is equivalent to the determinant of the coefficient matrix mentioned by @Mark44 .

So, yes, you did determine that there is a single point of intersection.

By the way: The method you were initially using in your Original Post, was correct, but your algebra contained errors.

I can follow up on that if you're interested.

Specter
SammyS said:
The product you calculated, known as the scalar triple product, is equivalent to the determinant of the coefficient matrix mentioned by @Mark44 .

So, yes, you did determine that there is a single point of intersection.

By the way: The method you were initially using in your Original Post, was correct, but your algebra contained errors.

I can follow up on that if you're interested.
Sure that would be great! Thanks!

Specter said:

The Attempt at a Solution

I have to pick a variable, use a pair of equations to eliminate the variable. Then I have to eliminate the same variable but with a different pair of equations. I tried doing this but I am not sure how correct it is.
=(4,1,-9)⋅[(1,2,3)×(2,-3,0)]
=(4,1,-9)⋅(9,6,-7)
=105

105≠0, the normal vectors are not coplanar so there is a single point of intersection.
The above is fine.

In the following, you choose to eliminate ##\ x \ ## in two equations, which is a fairly common approach. However, as you point out above, you can pick any variable and in the case of these three equations, equation [3] already has ##\ z \ ## eliminated. I'll say more on that later.
4x+y−9z=0 [1]
x+2y+3z=0 [2]
2x−3y = 5 [3] . . . (Note: I added 5 to get the constant on the right side. Less confusing that way. : Sammys ) .

4x+y−9z=0 [1]
4x+8y+12z=0 [4] : ( 4 × Eqn [2] to eliminate x. )
Subtract and the new equation is 7y+3z=0 [5].
Apparently, you intended to eliminate ##\ x\ ## by subtracting Eqn [1] from Eqn [4]. (As I stated in Post #2, the result you give as Eqn [5] is incorrect.)
Specifically, 12z − (−9z) → 21z, not 3z .

I find that I make fewer arithmetic errors, if I do the elimination by adding equations rather than subtracting them. So for your system of equations, get Eqn [4] by multiplying Eqn [2] by −4 .
Then add equations [1] and [4].

That gives: ##\ 0x - 7y - 21z = 0 \ ## for Eqn [5].
( Maybe divide this by −7 & use this instead. ##\ \ \ y + 3z = 0 \ ## , Eqn [5*]
Furthermore, this could also be written: ##\ 3z = -y \,.## Going even further, plug this back into Eqn[2] to get
##x + 2y + (-y) =0 ##
## x + y =0 \ ## ( Eqn [5**] )​
Again, more on this later .​
)​
Use a different pair of equations to eliminate x. I used equations [1] and [3].

4x+y−9z=0 [1]
4x−6z−10=0 [6] : ( 2 × Eqn [3] )
Subtract and the new equation is −5y−9z−10=0 [7]
Again, an arithmetic (or algebra) mistake due to incorrect subtracting.
That should have given: −7y + 9z − 10 = 0 for Eqn [7]

The new system with 2 eqns and 2 variables:

7y+3z=0
-5y-9z-10=0

If this is correct, I sort of know where to go from here. I would solve for y and z.
In writing the "system", it's customary to write a complete set of three equations. So, any of the original equations, Eqn [2] is recommended b/c the coefficient of x is 1 .

With the corrected equations, the system can be written as follows.
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ -7y + 9z &= 10 & \text{ [7*]} \nonumber
\end{align} \right. ##

Use elimination with Eqn's [5*] and [7*] to obtain an equation in which ## y ## is eliminated. Easily done by taking 7 times Eqn [5*] and adding Eqn [7*] to that.
You get ##\ 30z = 10 \ ## which simplifies to ##\ 3z = 1 \,. ##

The "system becomes:
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ 3z &= 1 & \text{ [8]} \nonumber
\end{align} \right. ##

The above system may be said to be in upper triangular form and the system can be solved by "back substitution".

This post is way long enough. I'll do the "later" stuff in another post. Basically, it involves using the observation that eliminating ##z## from an additional equation ( using elimination on Eqn's [1] &[2] ) gets you a result with much less computation.

Specter
SammyS said:
The above is fine.

In the following, you choose to eliminate ##\ x \ ## in two equations, which is a fairly common approach. However, as you point out above, you can pick any variable and in the case of these three equations, equation [3] already has ##\ z \ ## eliminated. I'll say more on that later. Apparently, you intended to eliminate ##\ x\ ## by subtracting Eqn [1] from Eqn [4]. (As I stated in Post #2, the result you give as Eqn [5] is incorrect.)
Specifically, 12z − (−9z) → 21z, not 3z .

I find that I make fewer arithmetic errors, if I do the elimination by adding equations rather than subtracting them. So for your system of equations, get Eqn [4] by multiplying Eqn [2] by −4 .
Then add equations [1] and [4].

That gives: ##\ 0x - 7y - 21z = 0 \ ## for Eqn [5].
( Maybe divide this by −7 & use this instead. ##\ \ \ y + 3z = 0 \ ## , Eqn [5*]
Furthermore, this could also be written: ##\ 3z = -y \,.## Going even further, plug this back into Eqn[2] to get
##x + 2y + (-y) =0 ##
## x + y =0 \ ## ( Eqn [5**] )​
Again, more on this later .​
)​

Again, an arithmetic (or algebra) mistake due to incorrect subtracting.
That should have given: −7y + 9z − 10 = 0 for Eqn [7]In writing the "system", it's customary to write a complete set of three equations. So, any of the original equations, Eqn [2] is recommended b/c the coefficient of x is 1 .

With the corrected equations, the system can be written as follows.
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ -7y + 9z &= 10 & \text{ [7*]} \nonumber
\end{align} \right. ##

Use elimination with Eqn's [5*] and [7*] to obtain an equation in which ## y ## is eliminated. Easily done by taking 7 times Eqn [5*] and adding Eqn [7*] to that.
You get ##\ 30z = 10 \ ## which simplifies to ##\ 3z = 1 \,. ##

The "system becomes:
## \left \lbrace \begin {align} x+2y+3z&=0 & \text{ [2]} \nonumber \\
\quad \quad \ y + 3z &= 0 & \text{ [5*]} \nonumber \\
\ \ \ 3z &= 1 & \text{ [8]} \nonumber
\end{align} \right. ##

The above system may be said to be in upper triangular form and the system can be solved by "back substitution".

This post is way long enough. I'll do the "later" stuff in another post. Basically, it involves using the observation that eliminating ##z## from an additional equation ( using elimination on Eqn's [1] &[2] ) gets you a result with much less computation.

Wow! ! I need to make it a habit to check my arithmetic over and over. Sometimes I feel overwhelmed and can't find the mistakes if that makes any sense. I appreciate you taking the time to type all of this out and explain it to me. Thank you, it helps a bunch!

Specter said:
I need to make it a habit to check my arithmetic over and over.
Well, yes, but if you get to a solution, you should always check that it is a solution to the original problem. We all make mistakes, especially in these kinds of problems where there are many arithmetic operations. If your solution checks, then that's a good sign that all of your work was correct.

Specter

1. What is a system of linear equations?

A system of linear equations is a set of two or more equations that contain two or more variables. The solution to the system is the set of values for the variables that satisfies all of the equations simultaneously.

2. How do you solve a system of linear equations?

There are several methods for solving a system of linear equations, including substitution, elimination, and graphing. The most common method is using the elimination method, where one variable is eliminated by adding or subtracting the equations to create a new equation with only one variable.

3. Can a system of linear equations have more than one solution?

Yes, a system of linear equations can have one, infinite, or no solutions. A system has one solution when the equations intersect at one point, infinite solutions when the equations are the same line, and no solutions when the equations are parallel lines.

4. What is the importance of solving a system of linear equations?

Solving a system of linear equations is important in many real-world applications, such as in engineering, economics, and physics. It allows us to find the relationship between different variables and make predictions or solve problems.

5. What is the difference between a consistent and inconsistent system of linear equations?

A consistent system of linear equations has at least one solution, while an inconsistent system has no solution. In other words, a consistent system has a point of intersection, while an inconsistent system has parallel lines that do not intersect.

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