Simultaneous Equations Algebratically

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fliss1992
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I need to solve the following equations algebraically..

9x - y = 12, 4y - 9x = 6

8x +3y = 6, 5x - 3y = 5

x +3y = 11, 2x +5y = 19

3x +3y = 9, y - x +5

3x + 2y = 21, 2x - 7 = 7

I've no idea how to solve these!?
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Doomlar
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You can do this one of two ways:

You can make the x terms equal in both the equations, I'll use your first set of equations to demonstrate this:


9x - y = 12

4y - 9x = 6

Multiply the second equation by -1 to get:


9x - 4y = -6

You can then proceed to subtract one from the other as follows:


9x - y = 12

- (9x - 4y = -6)

= 0x + 3y = 18

And then from there you can divide by 3 to find a value for y of 6, and then you can substitute that value back into either equation to find a value for x.



Alternatively, you can use the method of substitution, where you rearrange one equation to isolate a variable, and then substitute it into the other equation:


9x - y = 12

9x = 12 + y

y = 9x - 12

Then substitute into your other equation (4y - 9x = 6):


4(9x - 12) - 9x = 6

36x - 48 - 9x = 6

27x - 48 = 6

27x = 54

x = 2

You can then substitute this value back into either equation to find y.

Hope that helps!
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fliss1992
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(Original post by Doomlar)
You can do this one of two ways:

You can make the x terms equal in both the equations, I'll use your first set of equations to demonstrate this:


9x - y = 12

4y - 9x = 6

Multiply the second equation by -1 to get:


9x - 4y = -6

You can then proceed to subtract one from the other as follows:


9x - y = 12

- (9x - 4y = -6)

= 0x + 3y = 18

And then from there you can divide by 3 to find a value for y of 6, and then you can substitute that value back into either equation to find a value for x.



Alternatively, you can use the method of substitution, where you rearrange one equation to isolate a variable, and then substitute it into the other equation:


9x - y = 12

9x = 12 + y

y = 9x - 12

Then substitute into your other equation (4y - 9x = 6):


4(9x - 12) - 9x = 6

36x - 48 - 9x = 6

27x - 48 = 6

27x = 54

x = 2

You can then substitute this value back into either equation to find y.

Hope that helps!
I've not done maths before so I'm struggling to work it out?
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Doomlar
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(Original post by fliss1992)
I've not done maths before so I'm struggling to work it out?
The substitution method is quite difficult to get your head around at first, but it is the better method. Just try with the second set of equations, and post your working here. Follow these instructions:

  1. Rearrange one equation to get either x on its own, or y on its own, so you want either x = ay + b, or y = ax + b
  2. Now take your other equation ax + by = c and replace either x or y, depending on which you isolated in step 1, with what it equals in step 1.
  3. Expand the brackets and move the numbers around until its in a correct form.


Just have a go, and if you're stumped by anything I'll try and help!
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fliss1992
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I'm not sure how I'm rearranging things that's the beginning bit I'm stuck with, sounds stupid but I just don't get it.
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Doomlar
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(Original post by fliss1992)
I'm not sure how I'm rearranging things that's the beginning bit I'm stuck with, sounds stupid but I just don't get it.
Nah it doesn't sound stupid at all; some people get it some people don't!

If you have an equation, for example 2x + 3y = 5, if you want to do something to one side, then you have to also do it to the other, so if you want to remove the 3y on the left hand side then you take it away from the left hand side AND the right hand side.

So this means if you want to take away 3y then you would have 2x + 3y - 3y = 5 - 3y; and because 3y - 3y = 0, you have 2x = 5 - 3y, does this make sense?

If it does, then that's basically all there is to rearranging an equation

Then when you are dealing with your simultaneous equations, you need to rearrange one of the equations so you have x or y on the left hand side on its own.
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Doomlar
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Then when you are dealing with your simultaneous equations, you need to rearrange the equation so you have x or y on the left hand side on its own.
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