libertyd
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Does anyone know an easy way to do simultaneous equations? I've tried everything and just can't seem to get it right!

thanks
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chloe_32
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Rearrange one equation to make it x= or y= and then just sub it into the other equation. One of the unknowns won't be there, when you have the answer for one sub it back into a equation for the other unknown.
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edothero
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If you post an example question, or one you're specifically stuck on, we'd be able to put the method into context so you'd understand easier

But yes, as chloe_32 stated, I would rearrange either equation 1 or 2 and substitute it into the other.

For example;

If we have

Equation 1: 3x+5y-24=0 and
Equation 2: 5x+8y+36=0

If we take Equation 1 and rearrange for  x we get:

 3x=24-5y

\Rightarrow  x =  \dfrac{24-5y}{3}

Now if we substitute this into Equation 2

 \dfrac{5(24-5y)}{3}+8y+36=0

Now we can solve Equation 2 for  y

You would get  \dfrac{120-25y}{3}+8y+36=0

 120-25y+24y+108 =0

\Rightarrow  y=228

Now, let's substitute this y value into equation 1 to get the  x value.

 y=228

 3x+5y-24=0

 3x + 5(228)-24=0

\Rightarrow  x= -372

Therefore you have  x= -372 and  y=228

Hope this helps


P.S. I made the question up on the spot, don't be thrown off by the answers, concentrate on the method instead
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libertyd
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(Original post by chloe_32)
Rearrange one equation to make it x= or y= and then just sub it into the other equation. One of the unknowns won't be there, when you have the answer for one sub it back into a equation for the other unknown.
Thank you!!
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libertyd
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(Original post by edothero)
If you post an example question, or one you're specifically stuck on, we'd be able to put the method into context so you'd understand easier

But yes, as chloe_32 stated, I would rearrange either equation 1 or 2 and substitute it into the other.

For example;

If we have

Equation 1: 3x+5y-24=0 and
Equation 2: 5x+8y+36=0

If we take Equation 1 and rearrange for  x we get:

 3x=24-5y

\Rightarrow  x =  \dfrac{24-5y}{3}

Now if we substitute this into Equation 2

 \dfrac{5(24-5y)}{3}+8y+36=0

Now we can solve Equation 2 for  y

You would get  \dfrac{120-25y}{3}+8y+36=0

 120-25y+24y+108 =0

\Rightarrow  y=228

Now, let's substitute this y value into equation 1 to get the  x value.

 y=228

 3x+5y-24=0

 3x + 5(228)-24=0

\Rightarrow  x= -372

Therefore you have  x= -372 and  y=228

Hope this helps


P.S. I made the question up on the spot, don't be thrown off by the answers, concentrate on the method instead
Thank you so much!! In regards to the specific example how would you work out:

4x + 6y = 12
x + y = 1
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samb1234
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(Original post by libertyd)
Thank you so much!! In regards to the specific example how would you work out:

4x + 6y = 12
x + y = 1
Y=1-x
Substitute

Posted from TSR Mobile
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edothero
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(Original post by libertyd)
Thank you so much!! In regards to the specific example how would you work out:

4x + 6y = 12
x + y = 1
Well here we have a simple choice don't we.

You can start by rearranging equation 2 to get

 x = 1-y
or
 y= 1-x

To make it easier for yourself, always go for the simplest equation. No matter which equation you rearrange, the outcome will be the same.

Try it
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adam9317
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(Original post by libertyd)
Thank you so much!! In regards to the specific example how would you work out:

4x + 6y = 12
x + y = 1
So you would rearrange the second to be x=1-y or y=1-x its your choice.

Then sub it in

y=1-x
4x + 6(1-x) =12
And work from there

work out your X, then sub into original equation to get y
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