Solving a system of linear equations

[Specter]

Homework Statement

Sorry for all the posts lately. This should be the last one for a while.

Solve the following system of linear equations:

$\displaystyle 4x-y=10$

$\displaystyle x+y-3z=8$

$\displaystyle 3x-y+z=12$

Homework Equations

The Attempt at a Solution

[/B]
I started with finding if the planes intersect at a single point. $n \cdot (n_2 × n_3 )$ =31, so the planes intersect at a single point.

I've posted a question like this here before but I still don't understand it.

Eqn 1 only has 2 variables so I know I have to do something with that, but I am not sure what. I think If I can isolate a variable in equation 1 I can then use it to find the values for the other equations by substituing in, but I don't know where to start.

 

[PeroK]

$\displaystyle 4x-y=10$

Does this help you get started?

$y = 4x - 10$

 

[Ray Vickson]

Homework Statement

Sorry for all the posts lately. This should be the last one for a while.

Solve the following system of linear equations:

$\displaystyle 4x-y=10$

$\displaystyle x+y-3x=8$

$\displaystyle 3x-y+z=12$

You wrote the second equation as $x+y-3x = 8$. Do you really mean that, or should it be $x+y - 3z = 8?$

 

[Specter]

You wrote the second equation as $x+y-3x = 8$. Do you really mean that, or should it be $x+y - 3z = 8?$

Thank you, I fixed it.

 

[Specter]

Does this help you get started?

$y = 4x - 10$

Okay so now I have y, but how do I know which equation to plug y into?

This is what I tried:

Using eqn 2, solving for z:

$x+y-3z=8$

$x+4x-10-3z=8$

$5x-10-3z=8$

$5x-10-3z-5x=8-5x$

$-10-3z=8-5x$

$-10-3z+10=8-5x+10$

$-3z=-5x+18$

$\displaystyle z=\frac {-5x+18} {-3}$

If this is correct, can I just plug my y and z value into equation 1 to find x?

 

[PeroK]

Okay so now I have y, but how do I know which equation to plug y into?

This is what I tried:

Using eqn 2, solving for z:

$x+y-3z=8$

$x+4x-10-3z=8$

$5x-10-3z=8$

$5x-10-3z-5x=8-5x$

$-10-3z=8-5x$

$-10-3z+10=8-5x+10$

$-3z=-5x+18$

$\displaystyle z=\frac {-5x+18} {-3}$

If this is correct, can I just plug my y and z value into equation 1 to find x?

It's quicker and better to replace $y$ in both equations. But, what you have looks correct.

 

[Specter]

It's quicker and better to replace $y$ in both equations. But, what you have looks correct.

Thanks. So now I should sub in the same y value into eqn 3 and solve for x?

 

[PeroK]

Thanks. So now I should sub in the same y value into eqn 3 and solve for x?

It's quicker to sub $y$ into both equations, giving you two equations in $x$ and $z$. You then eliminate one of these to give you the other.

 

[Ray Vickson]

Okay so now I have y, but how do I know which equation to plug y into?

Plug it into each of the other two that you did not use to get that formula for $y$.

BTW: for setting up multi-line sets of equations in LaTeX, you get much better results using a displayed-equation setting using $ $ ... your equations ... $ $ (with no spaces between the two $s---I had to make such spaces in order to prevent LaTeX from trying to typeset an equation and choking on the results). Withing your displayed-equation environment you can use an "array" command. The format of an array is $\text{\begin{array}{positions}}$ first line \\ second line \\ ... last line \end{array}" You can have hundreds of line if you want. Exactly what goes into the "positions" option depends on the number of columns; for a single column it can be either "l" (no quotation marks) for "left justified", or "c" for "centered" or "r" for right-justified. If you want more than one column you must specify the position of each, so position = "rl" means that the array will have two columns, with the first right-justified and the second left-justified; within each line you need a column separator "&" to break up the line into two columns.

So, if I wanted to typeset your three equations with left justification I would use
$$ \begin{array}{l}
4x - y = 10 \\
x+y - 3z = 8 \\
3x - y + z = 12
\end{array} $$
However, it might look better if we used two columns:
$$ \begin{array}{lc}
4x - y &= 10 \\
x+y - 3z &= 8 \\
3x - y + z &= 12
\end{array} $$
You can right-click on each system and choose the "display math as tex" menu option to see the actual typed inputs.

 

[Specter]

It's quicker to sub $y$ into both equations, giving you two equations in $x$ and $z$. You then eliminate one of these to give you the other.

Plug it into each of the other two that you did not use to get that formula for $y$.

BTW: for setting up multi-line sets of equations in LaTeX, you get much better results using a displayed-equation setting using $ $ ... your equations ... $ $ (with no spaces between the two $s---I had to make such spaces in order to prevent LaTeX from trying to typeset an equation and choking on the results). Withing your displayed-equation environment you can use an "array" command. The format of an array is $\text{\begin{array}{positions}}$ first line \\ second line \\ ... last line \end{array}" You can have hundreds of line if you want. Exactly what goes into the "positions" option depends on the number of columns; for a single column it can be either "l" (no quotation marks) for "left justified", or "c" for "centered" or "r" for right-justified. If you want more than one column you must specify the position of each, so position = "rl" means that the array will have two columns, with the first right-justified and the second left-justified; within each line you need a column separator "&" to break up the line into two columns.

So, if I wanted to typeset your three equations with right justification I would use
$$ \begin{array}{l}
4x - y = 10 \\
x+y - 3z = 8 \\
3x - y + z = 12
\end{array} $$
However, it might look better if we used two columns:
$$ \begin{array}{lc}
4x - y &= 10 \\
x+y - 3z &= 8 \\
3x - y + z &= 12
\end{array} $$
You can right-click on each system and choose the "display math as tex" menu option to see the actual typed inputs.

Thank you.

So you say to plug the y value thaty I got from eqn 1 into eqn 2 and 3.

I've done equation 2 and got $z=\frac {-5x+18} {-3}$.

So now I plug y into equation 3? I would be solving for x, right?

Edit: Ord do I plug the two values that I already have, y and z, into equation 3 and solve for x like that?

 

[PeroK]

Thank you.

So you say to plug the y value thaty I got from eqn 1 into eqn 2 and 3.

I've done equation 2 and got $z=\frac {-5x+18} {-3}$.

So now I plug y into equation 3? I would be solving for x, right?

Yes, although simply:

$5x -3z = 18$

Was good enough in terms of what to do with equation (2). Just do the same sort of thing with equation (3).

 

[Specter]

Yes, although simply:

$5x -3z = 18$

Was good enough in terms of what to do with equation (2). Just do the same sort of thing with equation (3).

Would I plug both y and z into equation 3?

Like this:

$3x-4x-10-\frac {5x+18} {-3} =12$

 

[PeroK]

Would I plug both y and z into equation 3?

Like this:

$3x-4x-10-\frac {5x+18} {-3} =12$

You can do that!

 

[Specter]

You can do that!

Is there an easier way to do it? You said something about having 2 equations and then eliminating one but I don't understand what you mean. The way I am doing it I am so confused and the alegbra gets confusing with all of the fractions.

 

[PeroK]

You started off with:

$$ \begin{array}{l}
y = 4x - 10 \\
x+y - 3z = 8 \\
3x - y + z = 12
\end{array} $$
Which becomes:
$$ \begin{array}{l}
x+(4x -10) - 3z = 8 \\
3x - (4x -10) + z = 12
\end{array} $$
And simplifying gives:
$$ \begin{array}{l}
5x - 3z = 18 \\
z = x+ 2
\end{array} $$

Which is a simpler system of two equations in two variables.

 

[Specter]

You started off with:

$$ \begin{array}{l}
y = 4x - 10 \\
x+y - 3z = 8 \\
3x - y + z = 12
\end{array} $$
Which becomes:
$$ \begin{array}{l}
x+(4x -10) - 3z = 8 \\
3x - (4x -10) + z = 12
\end{array} $$
And simplifying gives:
$$ \begin{array}{l}
5x - 3z = 18 \\
z = x+ 2
\end{array} $$

Which is a simpler system of two equations in two variables.

Thank you this is starting to become a bit clearer. From here where do I go? I have 2 systems with 2 variables.

 

[PeroK]

Thank you this is starting to become a bit clearer. From here where do I go? I have 2 systems with 2 variables.

That's down to you now. Look at that last equation.

 

[Specter]

That's down to you now. Look at that last equation.

Ohhh!$5x-3(x+2)=18$

$2x-6=18$

$2x=24$

$x=12$$z=12+2 \\ z=14$

This is way easier than the other way I was doing it. So now can I plug these into either eqn 2 or eqn 3 to find y? Does it matter which one?

 

[PeroK]

Ohhh!

$5x-3(x+2)=18$
$2x-6=18$
$2x=24$
$x=12$

$z=12+2$
$z=14$

This is way easier than the other way I was doing it. So now can I plug these into either eqn 2 or eqn 3 to find y? Does it matter which one?

There's a simple way to do that as well. What's the simplest equation you have for $y$?

 

[Specter]

There's a simple way to do that as well. What's the simplest equation you have for $y$?

y=4x-10?

 

[PeroK]

y=4x-10?

What could be simpler!

 

[Specter]

What could be simpler!

Hmmm, I'm not sure. But would this work?

$(12)+y-3(14)=8$

$y=8-30$

$y=-22$

 

[PeroK]

Hmmm, I'm not sure. But would this work?

$(12)+y-3(14)=8$

$y=8-30$

$y=-22$

That's not right. $y = 4x -10$ has got to be the simplest way.

 

[Specter]

That's not right. $y = 4x -10$ has got to be the simplest way.

Oh. For some reason I thought I would need x too.

$y=4(12)-10$

$=38$

 

[PeroK]

Oh. For some reason I thought I would need x too.

$y=4(12)-10$

$=38$

You've got them all now. You could check that the three values you have solve one or all of the equations - to make sure you haven't made a mistake. Personally, I would always check at least one.

 

[Specter]

You've got them all now. You could check that the three values you have solve one or all of the equations - to make sure you haven't made a mistake. Personally, I would always check at least one.

Thank you for the help! I will check them on paper.

 

[Specter]

You've got them all now. You could check that the three values you have solve one or all of the equations - to make sure you haven't made a mistake. Personally, I would always check at least one.

Oh one more thing. This will work to solve any problem like this right? Will there always be one equation in the system that has one less variable than the others?

 

[PeroK]

Oh one more thing. This will work to solve any problem like this right? Will there always be one equation in the system that has one less variable than the others?

Having one equation in only two variables is a big help. Always focus on that first. But, sometimes you have all the variables in all the equations. There should be material in your course on how to tackle these.

 

[mathwonk]

there is a systematic way to solve all such problems just using multiplication and addition/subtraction, called "row reduction". here is an explanation on khan academy:

https://www.khanacademy.org/math/li...ination/v/matrices-reduced-row-echelon-form-1

it is well worth learning since it is one of the few solutions methods that always works.