How to Solve a System of Equations Using Elimination with Fractions
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- čas přidán 12. 11. 2014
- 👉Learn how to solve a system (of equations) by elimination. A system of equations is a set of equations which are collectively satisfied by one solution of the variables. The elimination method of solving a system of equations involves making the coefficient of one of the variables to be equal in all the equations that make up the system, in order to eliminate the variable using subtraction (or addition as the case may be). The coefficients can be made equal by multiplying the entire equation with a number which when multiplied to the coefficient of the variable to be eliminated gives the LCM of the coefficients of the variable to be eliminated in all the equations. Then we solve for the remaining variable(s).
After we have obtained the solution{s} for the remaining variable(s), we substitute the obtained solution into any of the original equations to obtain the solution of the eliminated variable.
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I'm commenting here to tell you that you've saved a student from failing a quiz tomorrow. Thank you very much sir,,,
Did u fail it?
Im here for fun
Did you pass?
But did you pass?
Its funny how ur now probably grade 11 now
Damn it nick! Sit down!
ooh man. my boy nick
I’ve learned more in a 4 minute and 12 second video than I have learned in 6 months thank you sm!!
I wasted my 30 min trying to solve such questions and now finallly I got it.
I was struggling with a question like this for an hour THANK YOU!!
I swear I’ve learned more from you than my “professor”. So thankful to have come across your channel!
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just happy to help
SAMEEE
Thanks so much! You're a lifesaver. I'm just reviewing algebra for geometry class, and I was always uncomfortable and confused with fractions.
Greetings Brian McLogan,
I hope this letter serves you well. There was a saying that went by going "Those with a shiny head shine light onto others." Before I was curious as to what that meant. But today my eyes have been opened to a new world. I would like to deeply thank you from the bottom of my heart for taking me out of the darkness with your head. It has helped me pass my math course this year.
Sincerely,
Chan
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Thank you, thank you, thank you! I was having trouble with this, but you explained it beautifully!
awesome! happy to help
Cramming before a test tomorrow. My boy Brian saving the day again, respect.
Sorry Brian I got a 62 lmao
@@quixbix what a mood xDD
I read it as brain not Brian, lol
Thank you very much, sir! This is a very clear, good and easy to understand explanation!
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Sir...on this last day of the semester for my only math class for college, your videos have helped me exponentially!!! Thank you so much!
I finally understand it a lot better, thanks . Hope everyone is well.
Your videos are so helpful!!! Only reason I pass my tests
Sir, you’re awesome. You’ve helped me so much.
Good one. Another way of doing it is to multiple the first equation by 2 since the aim of this method is to create at least two same terms(variables) that can cancel each other when added/"substrated
Honestly this helped a lot😭
I knew this method but i didnt know how to deal with fractional forms. Simple and clear explanation
yoooooh What a relief u helped me so much tnx i was really having a problem...thanks a lot
Thanks for your help....was really struggling with the fractions....God Bless
cheers!
Thank you so much it will help me for my Fraction Problems.
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I swear I would've failed my maths exam , if this man wouldnt have been here on yt
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Thank You so much Sir I couldn’t understand this in class . This is gonna help me so much in my finals
Thanks very much Mr McLogan i do this in a program called kumon and its not that charming, this helped a lot thank you so much
asian lol?
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I can't tell you how many exams I've taken confidently because I watch your videos. Thank you very much sir!
Thank you very much Sir, for making this video and saving me from getting bad marks.
amazing it helped me seriously a thousands of dozens in my paper and you know what I have saved my 54 marks out of 75 marks overall
happy to help
Thank you so so much!! I was able to complete my homework because of you!
You are very welcome Daniela
Crud. I just watched this video after my quiz. The fact that I'm the smartest boy in my math class.. I was unfamiliar with this so I changed everything into a decimal and just solved the system using substitution. The hard part was that 1/3y was in an equation. Well it was only a bonus question, but now I know how to do it. Thank you for your help.
good work, if you struggle, converting to decimals works, I prefer to keep fractions as it usually is simpler
Thanks so much I was struggling with how to do this with fractions
Thanks sir tomorrow is my exam and after watching this video now my concepts are clear you are the teaches well
You saved my life
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Cheers! Happy to be able to help you out Terry!
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Love the way you write the number '6'😜😜
And tell Nick to sit down in the class
Ik it’s 3 years later but THANK YOU SO MUCH!!
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The way he draws 6 triggers me...
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I know this is every comment and this vid is like 6 years old, but you just saved my grade thank you so much.
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no just record what I currently teach at my school
Thanks a lot professor, my professor goes too fast and skips steps. I'm always looking for help on CZcams.
happy to be there for you!
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Boom
So happy to help
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"Hey, sit down!"
Really helpful!!! Thanks!!
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happy to help
What if the number on the right of the equal sign was a fraction? Would you find the LCD of all 3?
Thank you so much sir this really helps 😇
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Thank you!
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If the variables are in denominator then how could equation be formed.??
What if he didn't multiply the bottom equation by -15 and work with the positive 15 could it work out
HUGE help
Very helpful!
THANKS!
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Thank u
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i'm 5 years late but thank you sooo muchhh i really want to use this method instead of substitution 😭 again, thank you so much professor 🤚🏼😔
Question! So if I didn't do it the way you did, where you changed those negatives in positives, would the following be incorrect with solving for y? :
(1/2x + 1/3y = 7) x6
(1/5x + 2/3y = -2) x15
---------------------------
3x + 2y = 42
- 3x - 10y = -30
---------------------------
-8y/8 = 8/8
-----------
y = -1
???
I do not believe you distributed the negative all the way through. Once you multiply you should have 3x +2y=42 and 3x+10y=-30 so when you subtract the two equations you obtain -8y=72
Jaiden Allen 42 - 30 = 12. You should get -8y=12, and after dividing both sides by -8, and then simplifying, we get that y = -(3/2).
I plugged y = -(3/2) back into 3x + 2y = 42, and we get that x = 15 (I will leave the solving to you).
The solution should be (15, -(3/2)). I even checked this answer on Wolfram Alpha for absolute verification.
According to question correct second equation is 1/5 x - 2/3 y = -2 where as you solved with 1/5 x +2/3 y = -2 that is the reason answer is not matching with u tube vdo
Brian McLogan teaching is excellent which made to solve similar harder problems easy.There will not be other answer of this Question as given in vdo.
@@brianmclogan Mr Logan, why was the -10y changed to +10y?
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Really Thank You So much Sir ❤
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+Jewoski Wingard happy to help!
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Thanks
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Happy to help Yong
Mr. McLogan, the solution should be x = 15 and y = -(3/2).
First, getting rid of the fractions, the LCD of the first equation is 6, and the LCD of the second equation is 15. Multiplying both sides of the first equation by 6 gives us that 3x + 2y = 42. Multiplying both sides of the second equation by 15 gives us that -3x - 10y = -30.
Now, after getting rid of the fractions in both equations, we see that our coefficients on x are opposites, which allows us to add the two new equations together. Doing so yields -8y = 12 for us, eliminating the variable x. Dividing both sides by -8, and then simplifying, we get that y = -(3/2).
To finish it off, using the equation 3x + 2y = 42, I plugged -(3/2) in for y, and solved for x.
3x + (2/1)*(-(3/2)) = 42
3x + (-3) = 42
3x - 3 = 42
Adding 3 to both sides gives us that 3x = 45, and then dividing both sides by 3, we get that x = 15.
Therefore, our solution is (15, -(3/2)).
I have checked this answer both by hand, as well as by using Wolfram Alpha.
I think you wrote the problem down wrong, the second equation after multiplying by 15 should not give you a -3x. I do not see any mistake in my work
Brian McLogan Hmmm... I could have sworn that I didn't write it incorrectly. Moreover, for the first term, (15/1)*-(1/5)*x = -(15/5)*x = -3x.
Ah... I see the problem, now. I did the problem in the thumbnail. Your coefficient on the x in the 2nd equation was a -(1/5) in the thumbnail, as opposed to the video, where it was written down on the board as a (1/5).
no worries, thank you for looking out though, I will have to fix that thumbnail
Ahh finally . A good teacher !
THANK YOU SOO MUCH 💓
Thanks so much
you are very welcome!
Thank you so much
Thanks sir
he did it the long way but it is helpful (:
Thx this helped alot
Damn it I love u Brian
Can you do captions for this video? Thanks.
From your solution, the equation two(2) is so different from its solutions....there was negative(-) sign attached to the 1/5 in equation 2 but the negative sign didn't affect your solution... please why