How would I solve this simultaneous equation?

So here is the equation.....

2x + 5y = 16
5x - 2y = 11

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.

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...

Reply 3

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OP

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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 :smile:

Reply 4

username2640523

8

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 :smile:

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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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 :smile:

No worries :smile:

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.

Reply 6

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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

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Sorry I don't understand this bit???

Reply 7

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Original post by RDKGames

No worries :smile:

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??

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)

(edited 7 years ago)

Reply 9

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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?

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.

Reply 11

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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?

(edited 7 years ago)

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?

Reply 13

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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?

Reply 14

B_9710

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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 ?

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.

Reply 16

133221333123111

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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.