Can anyone help me understand this past non-calculator maths question??? It's for my exam this afternoon!!! :L

  • 0 votes

Please can someone help me?!?! This a really hard question, and I did get it when I did my mock, but it took up a lot of my exam time. So please can someone help me so that I can work it out as quickly as possible in the exam?! PLEASE!!! :/
Here's the question:
'solve algebraically these simulataneous equations
3x - y = 1
5x + 3y = 4

x= , y= '
Really hope someone can help me, and really hope they can help as soon as possible. So I know what I'm doing for my exam this afternoon. Thank you!!! :)

Posted Mon 11th June, 2012 @ 10:55 by Paige Conlon

6 Answers

  • 1 vote

times the top equation by 3 so both have the same y value ( you can do it for y or x, but y is easier in this question) so it is 9x- 3y = 3

then because the y values have opposite signs ADD the equations ( if the signs are the same subtract)

so 9x - 3y = 3

+  5x + 3y = 4

=  14x = 7

so x = 7/14

x = 0.5

then substitute x into one of the equatons

3(0.5) + 3y = 4

1.5 + 3y = 4

3y = 4 - 1.5

3y = 2.5

y = 2.5/3

hope that helped :) good luck in the exam! Can you help me out with my question, I have the exam today as well

good luck

Answered Mon 11th June, 2012 @ 11:06 by Freya
  • 0 votes

First of all you need a match in the 2 equations, so you multiply the first one by 5 and the second by 3, giving you...

15x - 5y = 5

15x +9y = 12

Take the top equation away from the bottom one, leaving...

14y = 7

(divide by 14)

y = 0.5

Put y = 0.5 into the top equation (the original one)

3x - 0.5 = 1

(+ 0.5)

3x = 1.5

(divide by 3)

x = 0.5

Now you've worked out x and y, put them into the bottom equation (the original) to check that it works, and there you have it. :D

Answered Mon 11th June, 2012 @ 11:18 by Nick
  • 0 votes

Firstly, multiply each of the equations either by the integer in front of the 'x' or the 'y'

(3x - y = 1) × 3

(5x + 3y = 3) × 1

It is usually best to chose the letter with the lower values as this makes it less complicated, so I have chosen the 'y' values. Bear in mind that if there is no number in front of the letter, you should use '1'. Write this in if it makes it easier.

9x - 3y = 1

5x + 3y = 3

After you have done this you should be left with two equations, however either of the 'x' or 'y' values are the same. What you now do is work out whether you need to subtract the two equations or add them together in order to find your answer. In this case, as the signs are different, you add.

If the Signs are different add - If the signs are the same subtract

9x - 3y = 1               After doing this, I have got rid of the y values as they total 0.

5x + 3y = 3 +           Now, I am able to calculate what 'x' is using the equation '14x =4'

14x    = 4           After calculating that x is 4 ÷ 14 which equals 0.29 (2dp)

I am able to input the 'x' figure (0.29) into either of the starting equations to calculate 'y'

3x - y = 1         =          (3×0.29) - y = 1              =          8.7 - y = 1      

there for 'y' equals 8.7 - 1 = 7.7

I hope I have helped, however the numbers I have produced seem a bit dodgy :S please check whether they are correct or not using a mark scheme (if this is a past paper question) The method annotation is correct, so If  the answers are wrong, go through the calculation again, and find out where I went wrong. I'd give it another go but I must do some revision myself! :)

Anyway, at least you now know how to do these calculations. Good luck with your exam!

Answered Mon 11th June, 2012 @ 11:28 by Harry Barlow
  • 0 votes

It looks like mine was wrong!

Answered Mon 11th June, 2012 @ 11:29 by Harry Barlow
  • 0 votes

Thank you! xD that's really helped me out, and I totally understand it now!!! xD and yeah, sure. I'll see what I can do. :) what's the question???
Oh, and also, I just want to point out that you timesed the 0.5 which is the x by 3 when it should be timesed by 5. But it's fine cuz uv taught me what technique to do and luckily I spotted the small error. Thank you!!! xD

Answered Mon 11th June, 2012 @ 11:29 by Paige Conlon
  • 0 votes

Thank you Harry, that's fine! :) and Iv learnt how to do the equation so it fine, THANK YOU!!! xD and I hope I do well for my exam, and good luck for you exam as well! :)
We'll both have fun revising as well :P xD

Answered Mon 11th June, 2012 @ 11:50 by Paige Conlon