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I really don't understand them no matter what. I can do the easy ones but the rest are just too hard! 

Posted Sat 5th May, 2012 @ 15:21 by LP-FTW

4 Answers

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Can you help me on the easy ones too...? I sometimes still get confused on them. But thanks anyways.

Answered Sat 5th May, 2012 @ 15:53 by LP-FTW
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I prefer to solve simultaneous equations by substitution, so I'll do an example using that method.

I'll use the equations 2x+3y=6 and 4x-6y=-4

  • Rearrange one of the equations, so you get x=... or y=...
  • 2x+3y=6  =>  2x=6-3y  =>  x=3-(3y/2)
  • Substitute x into the other original equation
  • 4(3-(3y/2))-6y=-4
  • Expand the brackets 
  • 12-6y-6y=-4  =>  12-12y=-4
  • Solve it to find y
  • -12y=-16  =>  y=4/3
  • Substitute y into one of your original equations
  • 2x+3(4/3)=6  => 2x+4=6  => 2x=2  =>  x=1
  • So x=1 and y=4/3

Have a look at as it has worked examples for substitution and elimination methods and the layout is clearer than explaining it on here.

Answered Sat 5th May, 2012 @ 18:20 by Maleficent
Edited by Maleficent on Sat 5th May, 2012 @ 18:28
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Right, simultaneous equations can seem a doddle when you know what you're doing... It just takes a little time for them to click properly :)

There are different kinds of simultaneous equations, with the key being equating or substitution.

is when you have two equations with one variable in the main equation, and both equations equal the same thing (this could be anything, with the most common being the variable y for solving coordinates on graphs).

What you do is you put both equations equal to each other, eliminating the answer and allowing you to arrange the equation until you have just the variable on one side and all constants on the other.

If the original equations equalled a variable then you can substitute the value you just found back in to either of the original equations to find the other variable. (You can substitute this value back in to both original equations if you want, and if they come out with the same answer you can be pretty certain you got the right answer.)

The other method is substitution. This is usen when you have one equation with two variables in it and another equation involving the same variables, but equals one of the variables.

That might not make much sense, so an example of this would be with the equations:
x^2 + 3x - 5y = 15 and y = x + 4
(feel free to try solve that, but I just picked random numbers off the top of my head)

What you do is substitute the second equation in to the first so you have an equation with only one variable. You then rearrange this equation until you find the variable's value. Once you've done that, you can substitute that value in to the second equation to find the other variable's value.

With the above example this would work out as:
x^2 + 3x - 5(x+4) = 15 you rearrange this and find x
sub the value of x into y = x + 4 

Hope that helps anyone with a problem solving simultaneous equations, cause it took me ages to write! Haha.

Answered Sat 19th May, 2012 @ 16:58 by sammiecaine
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Simultaneous equations aren't that hard! :) Say you have these 2 equations:



You need to either make the y values or the x values the same in both equations. So let's go with making the x values the same...

2x+3y=13 (x3)

3x+2y=12 (x2)




Then you subtract the bottom equation from the top one, which gives you..


Then you divide it by 5

5y=15 (/5)


Now that you know y is 3, you can substitute this back into one of the equations to find x...

so  2x+3y=13 = 2x+(3*3)=13

 2x+9=13   (then if you -9)

2x=4 (and / by 2)


So there you go, x=2 and y=3 !

I know that is a simple one, but it would be useful if you could give an example of a harder one as I'm not really sure what you're struggling on! Hope I helped.

Answered Tue 22nd May, 2012 @ 19:10 by Hoosierette