Use back substitution to solve

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frctl

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Use back substitution to solve the following system of equations

Screen Shot 2019-09-08 at 4.56.27 PM.png

I am stuck since there are two unknowns in equation4.
Which operation can I take to solve this system?
 
Equation #4 has 2 unknowns, both of which appear in equation #3. If you solve equation #4 for x4, you can substitute that into #3 (reducing it down to 2 variables) and #2 (reducing it down to 3 variables). Continuing this process will eventually lead to #1 having one variable which can be solved. Then you start substituting back down the list to eventually find all 5 solutions.
 
frctl said:
Use back substitution to solve the following system of equations

View attachment 13522

I am stuck since there are two unknowns in equation4.
Which operation can I take to solve this system?
Click to expand...
You have 4 equations in 5 unknowns, so you can expect the solution to be parametrized, not a single point. Let [MATH]x_5 = t[/MATH], and work backward as R.M. just suggested.
 
x4 = 3x5
sub into equation 3
4x3 + (3x5) - 2x5 = 1
4x3 + x5 = 1

Which step should I take from here?
 
First the "bad" news: as stated above, your solution will not be a coordinate represented by single values ie. an ordered quadruple (3, 4, -1, 2, 5). You are looking at an infinite number of solutions, since you cannot solve each variable for a specific value.

The good news: you already have 2 solutions! You know that x5 equals itself, and x4 = 3x5. If you continue to solve each variable in terms of x5, you will have a complete solution!

Currently you have 4x3 + x5 = 1. Solve this for x3. Then sub what you have solved into the equation above that to solve x2 in terms of x5. With some persistence, you will eventually reach x1 in terms of x5, and then have a complete solution!
 
frctl said:
x4 = 3x5
sub into equation 3
4x3 + (3x5) - 2x5 = 1
4x3 + x5 = 1

Which step should I take from here?
Click to expand...
Here's an important question: What does your textbook say about "back substitution"? It presumably has told you what that means, and given an example, so that is the first place to go to find out what you are supposed to be doing.

Second, you don't appear to have read what I said. I know you have had problems involving parametric solutions, so you should be familiar with the idea of using something like [MATH]t[/MATH] as a parameter. Doing that will make what you are doing far less confusing than leaving [MATH]x_5[/MATH] in the equation.
 
Can we please see your result even if it is just for others to learn from. Thank!
 
Let's get you started, then.

The last equation is [MATH]x_4 - 3x_5 = 0[/MATH]. My suggestion was to start by letting [MATH]x_5 = t[/MATH], which will be the parameter for the solution.

That turns this equation into [MATH]x_4 - 3t = 0[/MATH]. Solve that for [MATH]x_4[/MATH], then put that into the next-to-last equation and solve that for [MATH]x_3[/MATH], and so on.
 
But [MATH]x_5[/MATH] is still [MATH]t[/MATH], right? Maybe you need to make that substitution in all the equations before proceeding.
 
I'm not sure how to do this since every x value is a different variable.
Some might be free variables although I'm not sure how to solve.

Oh you mean the following:
4x3 + (3t) - 2t = 1
4x3 + t = 1
 
Last edited:
Yes, that's right. Now, of course, solve for [MATH]x_3[/MATH].

Just keep working at it! Don't stop when you're unsure of something.

Often questions are answered when you do the next step. I think you're spending too much time asking questions, and not enough time learning by doing. If you don't trust yourself, trust the people who tell you what to do, and carry it out.

Maybe t is the free variable?
 
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