How to solve simultaneous linear equations … using algebra

February 15, 201348 Comments

How to solve simultaneous linear equations.. using algebra…

These videos show how to solve simultaneous linear equations in steps.

  • The first video should be relatively straightforward as it only deals with positive numbers.
  • The second is a little trickier (around level B) and involves dealing with a negative term.
  • The third video … shows more of a real application..

Click here for simultaneous linear equations quick test

Learning how to solve simultaneous linear equations can be important for applications in economics, such as working out the best price to sell a product. This is usually called ‘supply and demand.’

Imagine you make pencils: Simultaneous pencils

  • If you sell at a high price they’ll be less demand
    Simultaneous pencils
  • If you sell at a low price they’ll be too many and less profit

Simultaneous equations can be created to show how quantities sold vary with supply and demand. These can then be solved to show the best price to make sure you sell your pencils, and the demand continues.

Another example – a favourite in exams – is to use mobile phone contracts. Sometimes these are  given as a graph and there’s more about this in the next post. Although the question is usually two linear equations, and it asks you to pick the best value.

One of my favourite exam questions involves The Khans and The Smiths buying theatre tickets. Each family has got different numbers of adults and children … and you need to create a couple of simultaneous equations to work out the price of each ticket.

These kind of questions may be a little strange (why didn’t they just ring the box office?), but they do give an insight into how equations work. There are other examples such as arranging a meeting half way through a journey or working out the cost of  bank loans.

Please add a comment below with any more real life examples.

Watch the videos on YouTube:

How to solve simultaneous linear equations using algebra 

How to solve simultaneous linear equations using algebra 

How to solve simultaneous equation word problems 

 

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    Comments (48)

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    1. Cameron Richmond says:

      Very useful doing this in maths and I didn’t understand it till now and I
      can solve these in my up and coming maths exam

    2. davidymcmb says:

      Thanks for the help man!

    3. stephen barker says:

      ive nailed equations with minus numbers already for next year now

    4. BAT TT says:

      This is very helpful thank you

    5. Justin Beary says:

      Thank you so much I finally understand how to do it without making stupid
      mistakes wuhuuuuu!

    6. Annie Lim says:

      it is really blur

    7. mussa sherif says:

      Thanx

    8. killa killa says:

      Thanks you really helped I have a test tomorrow-hope I do well

    9. stephen barker says:

      i make 1 letter the same in both equations subtracted answers gets
      remaining letter

    10. mohamed abdelhalim says:

      thanks !!! really did help.could you solve this question for me please. A
      submarine can travel at 25 knots with the current and at 16 knots against
      it .Find the speed of the wind and the speed of the submarine in still
      water.

    11. SkizaKiza says:

      thankyou for the help!

    12. Simon Deacon says:

      Please like and leave a comment!

      Visit http://www.mathswrap.co.uk for real maths, tips and techniques.

    13. Satellite Dave says:

      Thank you very much!! Your video helped me a lot!

    14. Finlay McKee says:

      I have my GCSE’s this year and you have managed to teach me what my maths
      teacher has failed to. Thank you so much!

    15. Anita Cheung says:

      thanks

    16. Simon Deacon says:

      Hi – you need to plot both lines and see where they cross. It’s OK if you
      have an idea where they are likely to be on a graph… but it can take a
      long time to get that information. I’ll post a video on this and let you
      know. Simultaneous solving by using algebra is better and easier. All best S

    17. stephen barker says:

      a graph how would i do that

    18. Simon Deacon says:

      Hi – this is really the easiest way. You could solve by plotting a graph
      but it takes a while and isn’t always very accurate. Keep practicing and
      good luck!

    19. stephen barker says:

      whats the easyest way to solve these

    20. Simon Deacon says:

      Hi – Q2 needs a bit more explanation. If you email me through Maths Wrap I’ll send a solution. In the meantime Q1:

      You’ve got 2 equations J = 2L and J + L = 5L – 48.

      So, change both and you’ll get J-2L = 0 and J-4L = -48. Then take eq2 from eq1 and you should get -2L = -48. So L = 24.

      Put L=24 back into eq1 and J = 48.

      So Jan is 48 and Lisa is 24.

      I hope this helps and all best S

    21. mymindydiditong says:

      *two

    22. mymindydiditong says:

      Hi, im not sure how to solve these teo problems:
      1)Jan is twice as old as Lisa. The sum of their ages is 5 times Lisa’s age minus 48. How old are they now?
      2)John received changes worth $13 all in coins. He received 10 more dimes than nickles, and 22 more quarters than dimes. How many coins of each did he receive?

    23. Simon Deacon says:

      Hi Mustanser – glad you liked the video and thanks for the comment :-)
      The first equation is F = 3S . The second is a little more difficult. Imagine 10 years ago … at that stage the father would be F – 10 and the son would be S – 10.
      However the dad is 5 times older so F – 10 = 5 (S – 10). Now you’ve got two equations F = 3S (or F – 3S = 0) and F – 10 = 5S – 50 (or F – 5S = -40). Take equation 2. away from equation 1. You should end up with the son aged 20 and the dad aged 60. All best S

    24. mustanser hussain says:

      ago*

    25. mustanser hussain says:

      Hey,this was very useful indeed,thankyou.Can u give me the solution for this problem. A man is 3 times the age of his son.10 years aga he was five times the age of his son.Find their ages by finding the value of x.

    26. Simon Deacon says:

      Hi Glen – when I posted this it took out the new lines and doesn’t look as neat. I hope you can follow. If not please send your email address through mathswrap and I’ll send a reply.

    27. Simon Deacon says:

      Hi Glen – you’ve got two equations:

      C + Z = 25
      3.2C + 1.4Z = 62

      Multiply first by 3.2 (and leave second) so:
      3.2C + 3.2Z = 80
      3.2C + 1.4Z = 62

      Take second from first, so:
      1.8Z = 18

      Therefore Z = 10

      Then put back into
      C+ Z = 25
      C + 10 = 25

      So C = 15

      The alloy has 15kg of copper and 10kg of zinc.

      I hope this helps and thanks for the question.

      All best

      S

    28. Simon Deacon says:

      Hi Glen – you’ve got two equations:

      C + Z = 25
      3.2C + 1.4Z = 62

      Multiply first by 3.2 (and leave second) so:
      3.2C + 3.2Z = 80
      3.2C + 1.4Z = 62

      Take second from first, so:
      1.8Z = 18

      Therefore Z = 10

      Then put back into
      C+ Z = 25
      C + 10 = 25

      So C = 15

      The alloy has 15kg of copper and 10kg of zinc.

      I hope this helps and thanks for the question.

      All best

      S

    29. Glen Rossow says:

      The materials to make 25kg of an alloy of copper and zinc cost $62. If the copper costs $3.20/kg and the zinc costs $1.40/kg, find the composition of the alloy.

      How would i do that problem?

    30. Kim Khus says:

      it helps me thanks

    31. Simon Deacon says:

      Hi – generally yes, although you might need to change the equations a little. Thanks for the comment :-)

    32. zareensaba says:

      So Mr Simon .if we have two turms negative equation we add after we multiply .and with positive equations we subtract after the multiplication.right?!

    33. Devansh Sangoi says:

      oregata

    34. Simon Deacon says:

      Hi Ayisha – no, numbers change, although most examples tend to use ‘easier’ numbers. I’ll post a video with some harder questions and let you know when done. All best S :-)

    35. ayisha asuni says:

      anytime you have an equation do u always have to use the numbers 3 and 4

    36. Simon Deacon says:

      Hi – yep – went out and bought a better camera soon afterwards! Thanks for the comment and hope the vid was helpful :-)

    37. Ciara Mcsorley says:

      the only downside is that it isn’t in focus!!

    38. Simon Deacon says:

      Hi – this would be around B grade. If there was a ‘word’ problem that you needed to create the two equations, it would be an A / B.

    39. Georgia Pennant says:

      What grade is this?

    40. KellyReveiw says:

      thanks !!! really did help :)

    41. Simon Deacon says:

      Hi … hmm. There could be a way of getting a -7.5 if the width was also negative in your calculation. For most questions they would expect you to then convert to both positive numbers – and say the pool is 7.5m width. My email is on mathswrap – if you send me a copy (photo is fine) of your working – I’ll mark and email back. Hope this helps. S

    42. Nada Abdalla says:

      would it still be correct if we wrote that the width is equal to -7.5?

    43. David Taylor says:

      The step by step approach is just what students need and the inclusion of real life examples (why we learn this in the first place) is a great bonus.

    44. Simon Deacon says:

      :-) thanks

    45. lordmoonaz69 says:

      Cool, thanks for your videos :) you make it so easy to understand

    46. Simon Deacon says:

      Hi I’m a maths tutor and also run three first class learning centres.

    47. lordmoonaz69 says:

      Are you a maths teacher?

    48. Simon Deacon says:

      Simultaneous word problems coming soon!

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