Simultaneous Equations with Quadratics 

You all probably dont give a damn but does anybody know a tool to log back into an Instagram account..? I somehow forgot the login password. I would love any assistance you can offer me!

@Beckett Clyde instablaster ;)

Honestly 🫂

Emonur Rahman what kind of problem solving.?

@@MrAmw1 I think he means the one were say for example 3X+4y=7 Xsquared-4ysquared=9 Basically when the linear equation isn't equal to the quadratic Y value

Deacon how’s boozer

Feel free to email them: mathsexplained2@gmail.com. Then I can do a video

I have done one! ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hCXT_SbProo.html (sorry about the intro)

Maths Explained haha its fine. Thanks anyway.

Nice pfp

This video will hopefully be useful: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-1aq6N9wzP8s.html Skip to the end where I explain where the x values come from

@@maths.explained Ah thank you, I understand now:)

Quick? The video was 18 minutes

@@yy4770 for what I needed it was quick

It CAN be factorized, but it doesn't factorise nicely because there isn't a pair of whole numbers that add to make the number in front of the x and and multiply to make the number at the end

I cannot say. There are still two exams to go. They may not go up at all. If they do go up, they may go up 10 or 20 marks.

Ssddproblems.com

thanks for the suggestion. I'll make one soon

@@maths.explained thanks

Like my last video?

Not really. I have one question in particular, if it's not too much of a problem do you mind going through it on a video please.

Sure send it to my email: mathsexplained2@gmail.com

How is man gonna say "not really".The disrespect😂

My last example in the video involved two quadratics. It’s essentially the same process

For example, I found this: A company orders a certain number of bottles. 8 machines in a factory can make all the bottles in 5 days. All machines work at the same rate. For 2 days, only 4 machines are used. On the 3rd day, all 8 machines are used. Work out the total number of days to make the bottles.

+voidz (_voidz_) Sorry for the confusion, I just wrote what I saw in the question, but yes that is the correct answer and please do tell me how you got it :)

I'm no math teacher but I think I understand how to do these kinds of problems so let me explain: It takes 8 machines 5 days. to produce all of the bottles. In 1 day with 8 machines, 1/5 of the bottles are made because it takes 5 days to make all of them. So if 8 machines are being used for 2 days, 2/5 of the bottles are being made. However, since the question says that only 4 machines are used for 2 days, the productivity is halved so it will take twice as long. Therefore, 2/5 * 1/2 of the bottles = 2/10 of the bottles = 1/5 of the bottles will be made in the 2 days with 4 machines. 1/5 of the bottles have now been made in the first 2 days leaving 4/5 of the bottles to be made. 8 machines are being used from now on and we need to work out how many more days it will take to produce the final 4/5 of the bottles. It takes 8 machines 5 days to produce all of the bottles so it will take 8 machines 1 day to produce 1/5 of the bottles. Therefore it will take 8 machines 4 days to make 4/5 of the bottles (which are all of the remaining bottles). So after the first 2 days it will take 4 more to finish making all of the bottles. In total it will take 6 days to make all of the bottles.

I can't sleep and am very bored so I wrote out my working and I'll send it to you: imgur.com/gallery/rb1JlG4

+voidz (_voidz_) thank you so much for taking the time to do all of this, I really appreciate it! Make sure to get some rest, although it’s a bit ironic coming from me seeing as I’m awake now too :)

Yes haha I realised this when I was editing the video but didn't bother to change it. Thanks for pointing it out

Maths Explained thanks and your welcome I guess😊

Yes you can do it that way

Thanks, I never knew about your way and I actually think it's a lot easier so I'll use it if it comes up in the exam

X=+sqrt(7/3) and -sqrt(7/3)

Check it carefully, there is no middle “x” term

@@maths.explained sorry not sure where the 7 and 3 are coming from my quadratic equation was something like -12+sqrt144-4(5-41)/10 because b is 12 a is 5 and c is -41 but I’m getting like a huge nunber of 3424. I’m not very good at these so it might be my understand of what you’ve done

Oh I thought you meant the last question in the video

​@@jordanwhysall3005 make sure you put a multiplication between the 5 and -41, it looks like you have written (5-41) when it should be (5*-41). Also, make sure you enter it correctly with brackets around the numerator and square root.

Yes, it doesn't matter which one you rearrange and substitute in, it's just it was easier to make it y= in this example

Grade 4 is about 50%

No I thought it may come up in paper 2 or 3.

oh right thank you, I'll revise this tonight!!

That’s a good point actually. I should go through how to solve these with a scientific calculator

Does it work with the normal calculators? or just those special casio ones?

It should work with the usual scientific calculators. I think it’s called Casio FX83

Vonex it only works with those special ones , not the ones everyone uses for gcse

thats my point yeah, on how its unfair