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How would I solve this simultaneous equation?

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1. So here is the equation.....

2x + 5y = 16
5x - 2y = 11
2. make one equation equal to x and sub that equation of x into the other equation, then sub the value you get for y into either equation and solve for x.
3. (Original post by JackT2000)
So here is the equation.....

2x + 5y = 16
5x - 2y = 11
By multiplying 2x+5y=16 by one factor, and 5x-2y=11 by another in order to balance the coefficients of either x or y in both equations...
4. (Original post by RDKGames)
By multiplying 2x+5y=16 by one factor, and 5x-2y=11 by another in order to balance the coefficients of either x or y in both equations...
Ok so I now have

10x + 25y = 80
10x - 4y = 22

Btw what numbers do I use to organise these two is it 3 & 4? Sorry I'm so OCD
5. (Original post by JackT2000)
Ok so I now have

10x + 25y = 80
10x - 4y = 22

Btw what numbers do I use to organise these two is it 3 & 4? Sorry I'm so OCD
Now you have
10x=80-25y
10x=22+4y

Now you can work out y and then as a result work out x

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6. (Original post by JackT2000)
Ok so I now have

10x + 25y = 80
10x - 4y = 22

Btw what numbers do I use to organise these two is it 3 & 4? Sorry I'm so OCD
No worries

What do you mean by organising? Do you mean labelling them? You can just call the first equation (1) and the second (2)

Now you can perform (1)-(2) or (2)-(1) whichever one you prefer, this will leave you with only one variable for which you can solve.
7. (Original post by HFancy1997)
Now you have
10x=80-25y
10x=22+4y

Now you can work out y and then as a result work out x

Posted from TSR Mobile
Sorry I don't understand this bit???
8. (Original post by RDKGames)
No worries

What do you mean by organising? Do you mean labelling them? You can just call the first equation (1) and the second (2)

Now you can perform (1)-(2) or (2)-(1) whichever one you prefer, this will leave you with only one variable for which you can solve.
So how would I do (1) - (2)

Take everything in the top equation from the bottom??
9. The other way round, you'd be taking everything in the bottom FROM the top.

What you said would mean that you are doing (2) - (1)
10. (Original post by RDKGames)
The other way round, you'd be taking everything in the bottom FROM the top.

What you said would mean that you are doing (2) - (1)
Sorry for so many posts but does it matter what way round I do it either 2 - 1 or 1 - 2?
11. (Original post by JackT2000)
Sorry for so many posts but does it matter what way round I do it either 2 - 1 or 1 - 2?
As I've mentioned, you can do it either way you prefer since you've already balanced one of the coefficients for both of the equations and they are both positive.
12. (Original post by RDKGames)
As I've mentioned, you can do it either way you prefer since you've already balanced one of the coefficients for both of the equations and they are both positive.
Ok so I decided to take away (1) from (2)

10x + 25y = 80 (1)
10x - 4y = 22 (2)

10x - 4y = 22 (3)
10x + 25y = 80 (4)

(4) - (3) = x + 29y = 58 ???

And is it ok to label them like this does it make sense?
13. (Original post by JackT2000)
Ok so I decided to take away (1) from (2)

10x + 25y = 80 (1)
10x - 4y = 22 (2)

10x - 4y = 22 (3)
10x + 25y = 80 (4)

(4) - (3) = x + 29y = 58 ???

And is it ok to label them like this does it make sense?
You are correct; however, are you sure about that x there?
14. (Original post by RDKGames)
You are correct; however, are you sure about that x there?
10x - 10x = 0 so no x is needed??

So is it just + 29y = 58 ?

Can you guide me through what to do next so I don't have to keep asking please thx?
15. (Original post by JackT2000)
10x - 10x = 0 so no x is needed??

So is it just + 29y = 58 ?

Can you guide me through what to do next so I don't have to keep asking please thx?
y=2 ?
16. (Original post by JackT2000)
10x - 10x = 0 so no x is needed??

So is it just + 29y = 58 ?

Can you guide me through what to do next so I don't have to keep asking please thx?
Solve for y. Substitute that value into any other equation, solve for x and once you have both you are done. Simple as that.
17. (Original post by JackT2000)
10x - 10x = 0 so no x is needed??

So is it just + 29y = 58 ?

Can you guide me through what to do next so I don't have to keep asking please thx?
You've asked a very similar question before - you can apply the same logic.

You will keep asking if you don't do the steps for yourself. That is the only way anything is properly learnt in maths.
18. (Original post by RDKGames)
Solve for y. Substitute that value into any other equation, solve for x and once you have both you are done. Simple as that.
(//29) 29y = 58 (/29)

y = 2
19. (Original post by JackT2000)
(//29) 29y = 58 (/29)

y = 2
Correct. Now go find x.
20. (Original post by RDKGames)
Correct. Now go find x.

2x + 5y = 16 (1)

Check
2x + 5(2) = 16
2x + 10 = 16
- 10 -10

2x = 6
/2 / 2

x = 3

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