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Calcmathlete
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Note: This is not a question.
Tutorial for Solving Systems of Linear Equations (Two Variables)
 2 years ago
 2 years ago
Calcmathlete Group Title
Note: This is not a question. Tutorial for Solving Systems of Linear Equations (Two Variables)
 2 years ago
 2 years ago

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Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
There are several ways to solve systems of linear equations. The most common methods are by graphing, elimination, and substitution. Let's start off with one of the most basic methods, graphing.  Graphing Method  2x + y = 3 3x + 2y = 6 In order to solve this system using the graphing method, we first have to change the two equations into slopeintercept form. 2x + y = 3 > y = 2x + 3 3x + y = 7 > y = 3x + 7 Then, we graph these two lines. (Attached Below) The solution set of a system of linear equations when graphing is actually the point at which the two lines intersect. So by graphing the two lines, we can obviously see that the solution set of this problem is (4, 5).  Elimination Method  The concept of elimination revolves around the concept of adding two equations. Using an example, let's see what happens when you add equations together. 2x + y = 3 3x + 2y = 6  5x + 3y = 9 Do you see how this works? Now, let's say that the orientation of these two equations were different. What would you do then? 2x + y = 3 6  3x = 2y If this situation occurs, you have to rearrange it in a way that the form of the equations match. For example, if you have one in standard form, you have to algebraically return the other equation to the same form. 2x + y = 3 6  3x = 2y > 6 = 3x + 2y > 3x + 2y = 6 Now that the equations are in the same form, we can begin to solve. However, let's start with a simpler system to demonstrate the concept. 2x  y = 5 3x + y = 5 The process of elimination involves adding equations in a way that one of the unknown variables disappears. In this first example, let's see what happens when we simply add them right away. 2x  y = 5 3x + y = 5  5x + 0 = 10 Using the Identity Property of Addition, we can simply get rid of the 0. 5x = 10 Now it's all algebra. 5x = 10 x = 10/5 x = 2 Now we have one of the variables. You may be thinking "what about the other variable?". Well, in order to solve for the variable, all you have to do is plug what you have for this variable back into the equation like so: 3x + y = 5 3(2) + y = 5 6 + y = 5 y = 1 It doesn't matter which equation we plug it into because you will get the same result. Therefore, our solution set is (2, 1). Now, this seems very simple, but what would you do if you had something like this: 2x + y = 3 3x + 2y = 7 Let's see what happens when you simply add the two equations. 2x + y = 3 3x + 2y = 7  5x + 3y = 10 It doesn't look pretty does it? In fact, we can't even solve for the variables this way. So what do we do? Well, we do something called manipulating the equation. We multiply one of the equations in the system by a number so that we can cancel one of the variables when we add them. 2x + y = 3 3x + 2y = 7 2(2x + y = 3) 3x + 2y = 7 4x  2y = 6 3x + 2y = 7 Now, if we were to add these two equations, we would be able to cancel one of the variables. 4x  2y = 6 3x + 2y = 7  x + 0 = 1 At this point, using the information above, we can once again find the solution set. x = 1 x = 1 2x + y = 3 2(1) + y = 3 2 + y = 3 y = 5 Solution Set: (1, 5) If you're wondering why this works, allow me to tell you. By multiplying both sides of the equation by a number, you are actually balancing an equation. If we were to only multiply one side, then the values of x and y would change. This is a basic rule of algebra: "What you do to one side of the equation, you must also do to the other" Finally, to demonstrate a mastery of the elimination method, let's try solving this system of linear of equations. 4x – 3y = 25 –3x + 8y = 10 Now right off the bat, you will notice that no matter what you multiply one of the equations by, you can't get to a place where one of the terms is canceled. So, we have to multiply both equations in order to solve this. 4x – 3y = 25 –3x + 8y = 10 8(4x – 3y = 25) 3(–3x + 8y = 10) 32x  24y = 200 9x + 24y = 30 Now we can solve this system. 32x  24y = 200 9x + 24y = 30  23x + 0 = 230 23x = 230 x = 10 4x – 3y = 25 4(10) – 3y = 25 40  3y = 25 3y = 15 y = 5 Solution Set: (x, y) = (10, 5)  Substitution Method  The substitution method is a method that is based off a rule of algebra. Basically, it's similar to using a formula. You substitute values into the formula. The same thing applies here. Let's say we have the following system: y = 2x + 3 5x + 2y = 24 Before we turn over to elimination, let's try using our new friend substitution. It clearly says that y is equal to 2x + 3. This means that we can substitute 2x + 3 in for y. 5x + 2y = 24 5x + 2(2x + 3) = 24 5x + 4x + 6 = 24 Now that we have eliminated a variable, we can solve the system. 9x + 6 = 24 9x = 18 x = 2 Now let's plug this back into one of the equations. I prefer to plug it into the equation that has a variable isolated. y = 2x + 3 y = 2(2) + 3 y = 4 + 3 y = 7 Therefore, the solution set is (x, y) = (2, 7). Now, if we don't have a variable already isolated, we have to algebraically solve for a variable. It is very similar to converting standard form to slopeintercept form when graphing.  Special Cases  There are two special cases when dealing with systems of equations. Let's say we have the given system. 2x + y = 5 4x + 2y = 10 Let's try solving this. 2(2x + y = 5) 4x + 2y = 10 4x  2y = 10 4x + 2y = 10  0 + 0 = 0 0 = 0 Uhoh. Both of the variables disappeared! This is what's called an identity case. An identity case occurs when both variables disappear and both sides have the same value. This means that 6 = 6 is also an identity case. Now, what does an identity case mean? It means that it's the same line. Since it's the same line, there are infinitely many solutions because the lines intersect in every way possible. This is also called an infinitely many solutions case. Now, let's take a look at the second special case. 2x + y = 9 4x + 2y = 4 2(2x + y = 9) 4x + 2y = 4 4x – 2y = 18 4x + 2y = 4  0 + 0 = 14 0 = 14 Once again, both variables disappeared. This is called a case of no solution. This occurs when the two lines are parallel. Since parallel lines never intersect, there is no solution.
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
So Long! Any revisions?
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.1
GaussianJordan elimination.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.1
Cramer's rule
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Oh ok. I didn't know that that was the official name ;)
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.1
Nothose are excluded.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.1
You should include those.
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
I knew about Cramer's Rule, but I wanted to keep this on an Algebra I level. If i added Cramer's Rule and stuff, it would be borderline Algebra II since most classes don't teach Cramer's Rule until Algebra II if at all.
 2 years ago

Limitless Group TitleBest ResponseYou've already chosen the best response.1
Who cares if it's Algebra I or Algebra II? Knowledge is knowledge. Cramer's rule involves determinants. GaussianJordan elimination involves matrices and row operations. Neither is that hard.
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
I suppose. THe tutorial was just so long and it would've taken a while teaching determinants and all of that. Maybe next time. Thanks for the suggestion though :)
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
it looked great .. keep it up calcmathlete. gr8
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
There is one more method... Can I suggest??
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
@waterineyes Sure @mathslover Thank you :)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
@Calcmathlete when two lines are parallel then they have infinitely many solutions and not no solution..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Really? I thought that since parallel lines never intersect and a solution is an intersection, there would be no solution?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Yes.. That is why they are parallel lines and can have many infinitely solutions.. I will give the examples of both..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
What can you say about these two lines having equations: 2x + 3y = 9 And, 4x + 6y = 18 Now, according to you they have no solution??
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
I thought so? Parallel lines never intersect no? And since they never intersect, there are no solutions? I thought it was infinitely many solutions when it was the same line.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
I will show how how you can tell about the Nature Of Roots..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Wait, I just realized. The equations you gave are the same line right? That would be an infinitely many solutions. Oops
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
NATURE OF ROOTS: \[ax_1 + by_1 + c_1 = 0\] \[ax_2 + by_2 + c_2 = 0\] 1) UNIQUE SOLUTION: \[\huge \color{blue}{\frac{a_1}{a_2} \ne \frac{b_1}{b_2}}\] 2) INFINITELY MANY SOLUTIONS: \[\huge \color{violet}{\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}\] 3) NO SOLUTION: \[\huge \color{green}{\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}}\]
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Now I will tell about that other Method...
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Alright. Thanks! I learn something new everyday! :) So to make sure I have everything completely understood. Would this system be infinitely many or no solution? x + y = 3 2x + 2y = 4
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
No solution..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Parallel Lines + Overlapping Lines = Infinitely Many Solutions.. Parallel Lines + Non Overlapping Lines = No Solution..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Ok. That's what I thought. But using that logic, my original example in my tutorial: 2x + y = 9 4x + 2y = 4 I said that it was no solution. It was in actuality infinitely many solutions?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
It is having no solution..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Then what was the original problem? Oh wait. You're original comment says that you were just gonna show another method. My bad :)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Yes I will.. Do you know the name of this method??
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
The name of this method is : SOLVING SIMULTANEOUS EQUATIONS BY CROSS MULTIPLICATION... Any Idea if you have heard this before??
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
No? I've only learned Graphing, Substitution, Elimination, and Cramer's Rule. Also I'm starting to learn Gaussian Elimination. I'm interested now :)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Can you wait?? It will take time to explain to write and to give an example too.. May be I have to draw it too to explain more..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Alright. np :)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
SOLVING SIMULTANEOUS EQUATIONS BY CROSS MULTIPLICATION METHOD: This method is simple one and not too hard but one must know this method.. Firstly we have to write the given equations in Standard Form that is: \[\huge ax_1 + by_1 + c_1 = 0\] \[\huge ax_2 + by_2 + c_2 = 0\] Then, make the following fractions: \[\frac{x}{(b_1c_2  b_2c_1)} = \frac{y}{(c_1a_2  c_2a_1)} = \frac{1}{(a_1b_2  a_2b_1)}\] Once you get this, firstly to find x, equate First Fraction with the Last Fraction.. And to get y, equate second and the Last Fraction..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
I will give you an example that will make sense of this method..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
I am taking your Elimination Method Example @Calcmathlete .. 2x  y = 5 3x + 2y = 5 First Step : Write them in Standard Form: 2x  y  5 = 0 3x + y 5 = 0 Second Step: Identify the values of \(a_1, b_1, c_1, a_2, b_2, c_2\).. Third Step: Plug in the values in the formula: \[\frac{x}{(1)(5)  (1)(5)} = \frac{y}{(5)(3)  (5)(2)} = \frac{1}{(2)(1)  (3)(1)}\] \[\frac{x}{10} = \frac{y}{5} = \frac{1}{5}\] Third Step: To find x, equate: \[\frac{x}{10} = \frac{1}{5}\] So, \(x = 2\) To get y, equate: \[\frac{y}{5} = \frac{1}{5}\] \[y = 1\] So, by Solving with Cross Multiplication Method, we get: \((x, y) = (2, 1)\)...
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Sorry there will not come 2y it is simply y in the second equation of first step..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
Second equation that I have taken is: 3x + y = 5 and not: 3x + 2y = 5 Please note this point.. Solution is correct and is according to 3x + y = 5
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Oh. Ok. I see where the name of cross multiplication comes from. I guess it's just a matter of remembering the order/pairings. Is there a specific time where this would be quicker than the other methods like how you know when to use substitution and elimination?
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
We always use Substitution Method when in any of the equations given, The coefficient of either y or x is 1 it becomes easy to solve.. Like: 3x + 2y = 13 x + y = 5 Now just find x or y from second equation and put it in first equation you have y = 2 and x = 3.. For Elimination: If in the given equations: If you can get the same coefficient of x or y by multiplying something than you will use Elimination: Like: 5x + 4y = 18 10x + 7y = 34 See here when you multiply first equation by 2 then coefficient of x are becoming equal.. 10x + 8y + 36 10x + 7y = 34 Subtract them: y = 2 and so, x = 2.. You can use cross multiplication method also if you are good in multiplication and signs.. Because we always do mistakes with the signs...
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
I have one pictorial representation of cross multiplication method: dw:1341466088360:dw
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
dw:1341466207150:dw
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
I get it! :D Parts of it remind me a bit of Cramer's Rule and parts of it remind me of the Law of Sines.
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
Once again, thank you!
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
the author of kramer's rule...
 2 years ago

agentx5 Group TitleBest ResponseYou've already chosen the best response.0
@dpaInc lol This is an interesting read btw, lots of good information. Want to put it all together into one collaborative document in a new topic/thread @Calcmathlete ? I want to give you a medal to afterall :D (gave the previous one to @waterineyes for showing me something I didn't know)
 2 years ago

zepp Group TitleBest ResponseYou've already chosen the best response.0
Ishaan gave me an advice when I made my first tutorial, he told me to keep it short, a post that's too long would discourage some people to read through it, even if it's useful;, but good job! :D
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
@agentx5 But you know now that's the best thing.. If anyone has knowledge lets other light their candle on it..
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.10
I'll probably do that later when I get all of the other methods that I missed. In any case, it's long...lol @zepp Yeah. I noticed. Maybe I should pick a topic not as long like split this into just elimination method since that in itself is pretty long.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.2
There is nothing to do with Long and short... If someone want to increase his or her knowledge and has interest then he or she will read the whole.. But the important thing is you should write the words or give the tutorial in a way that can be easily understandable by the reader... Reader gets bored only when he or she is not able to understand what you have written..
 2 years ago

Eyad Group TitleBest ResponseYou've already chosen the best response.0
Good Job,@Calcmathlete .
 2 years ago

Trexy Group TitleBest ResponseYou've already chosen the best response.0
Thanks @Calcmathlete
 2 years ago
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