# Pre-IB Mathl: Simultaneous Equations.

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Hi. I'm having problems with this simultaneous equation. I would really appreciate your help.

2x + 3y = 17

7x - 5y = 19

2x + 3y = 17

3y = -2x + 17

y = (-2x + 17)/3

Hence,

7x - 5(-2x + 17)/3 = 19

7x + (10x - 85)/3 = 19

And this is where I get stuck.

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hey im in pre-ib math as well enterin sl next year. Well here goes where u ended off.

7x + 10/3x -85/3= 19

31/3x= 142/3

x= 142/31

sub into eq'1

2(142/31) + 3y=17

284/31 + 3y= 17

y= 81/31

and u can check ur answers by putting it in one of the equations.

derefore x= 142/31 and y= 81/31

cya hope this helped.

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Thanks for making an effort in helping me!

However, I just don't seem to get it.

I mean, how do you get from here...

7x + 10/3x -85/3= 19

...to here?

31/3x= 142/3

x= 142/31

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You are making it too complicated.

2x + 3y = 17 --> Call this equation (1)

7x - 5y = 19 --> Call this equation (2)

(1) x 5 = [10x + 15y = 85] --> Call this equation (3)

(2) x 3 = [21x - 15y = 57] --> Call this equation (4)

(You want the equations to have a same component (in this case the 15y) so that you can eliminate it)

(3) + (4) eliminates the y part (because 15y + (-15y) = 0y)

So (3) + (4) gives you 31x = 142 which means x = 142/31 = 4.58.... (but I'd leave it as a fraction to be exact)

Then substitute x into one of the equations, say (1), you have

(2)(142/31) + 3y = 17

3y = 17 - (2)(142/31)

3y = 7 26/31

y = 2 19/31 = 2.61 (3sf)

So your results are y = 2.61 (3sf) and x = 4.58 (3sf)

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What you're much better off doing is making a matrix equation out of it. You'll need to learn matrices anyway so may as well start now

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Yeah. Doing it as A Matrix would sovlve t very quickly.

Enter into you GDC:

Post by: Aboo

|2 3 |

|7 -5 |

|17|

|19|

Then since you cant divide matrices you have to multiply by the inverse, soenter into you GDC:

Post by: Aboo^-1 *

You haveto put A fst becauase the 1 in the 2 x 1 of the solutions matrix doesnt match the first 2 in the 2 x 2 of the coeficiants matrix.

You should get

x= 142/31

y=81/31

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or you could solve this graphically using your gdc.Solve for y and enter that into your calculator and look for the point they intersect.

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hey friend its easy to solve simultaneous equations and every one has his own way in solving it

2x + 3y = 17 equation 1

7x - 5y = 19 equation 2

i use the substitution method, substitute the second equation in the first one but befre that make x or y the subject of the formula

equation 2 (19 + 5y) / 7 = x

equation 1 2(19 + 5y)/7 +3y = 17

solve equation one to find y

equation 1 (38 + 10y)/7 +3y =17

equation 1 (38 + 10y)/7 =17 - 3y

then just cross multiply

equation 1 119 -21y = 38 + 10y

equation 1 31y = 81

now use y to find x and substitute in any of the equations lets take the first one

1. 2x + 3y = 17

equation 1 2x + 3(2.6129) =17

equation 1 2x = 9.1613

Edited by jojo

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What you're much better off doing is making a matrix equation out of it. You'll need to learn matrices anyway so may as well start now

There's _no_ point in making a matrix out of a simple simultaneous equation unless you have to! It takes a lot longer to do it especially without a calculator, on top of being confusing! So I suggest solving it the 'classical' way!

Oh and in your answers go for 3 significant figures instead of 2 decimal places because 3sf is what IB wants.

Yeah. Doing it as A Matrix would sovlve t very quickly.

Enter into you GDC:

Post by: Aboo

|2 3 |

|7 -5 |

|17|

|19|

Then since you cant divide matrices you have to multiply by the inverse, soenter into you GDC:

Post by: Aboo^-1 *

You haveto put A fst becauase the 1 in the 2 x 1 of the solutions matrix doesnt match the first 2 in the 2 x 2 of the coeficiants matrix.

You should get

x= 142/31

y=81/31

If you do this, at least IB wants you to show these steps first

Post by: Aboo*[X]=

[A-1]Post by: Aboo[X]=[A-1]

[X]=[A-1]

Otherwise you won't get full method points.

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Sorru, but I am going to have ot agree with lilybean.

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Sorru, but I am going to have ot agree with lilybean.

well my explanation wasnt detailed. but its funny how i posted my solution and we're learning how this exactly works with matrices. Anyways,

yea you could do it how the other guy did it (lilybean) its called elimination. The way i did it was called substitution. In this case elimination would

have been easier but im really used to substitution thats why i did it my way. anyways ur teacher SHOULD HAVE taught u both ways. If you still dont

understand and like my way (substitution) ill repost it with detail. sorry for the lack of detail in my first one.

edit- learn elimination because it is useful cause u dont have to deal with too many nasty fractions. plus my solution was with the help of a calculator. god knows

if i can do it normally. cya

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for most problems substitution > elimination, since it might be hard to rewrite them in a form that can be eliminated. So you definitely want to learn substitution.

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If you have a Casio calculator you don't even need to do anything, you just stick it in the GDC and it does everything for you . But I suppose most of you would have TI.

Yeah you should know both elimination and substitution. I guess I just learnt elimination first so I'm used to it. Only do matrix when the question asks you do do matrix. Matrix looks easy but it takes longer.

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If you have a Casio calculator you don't even need to do anything, you just stick it in the GDC and it does everything for you . But I suppose most of you would have TI.

Yeah you should know both elimination and substitution. I guess I just learnt elimination first so I'm used to it. Only do matrix when the question asks you do do matrix. Matrix looks easy but it takes longer.

Well it's simple with a TI as well, but you are sometimes required to solve this algebraically, like in SL/HL Paper 1. And I agree what you say about Matrices, however, IB SL tests usually ask for matrix solution, at least according to my teacher.

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Yeah. But it is very helpfull later on when you start dealing with larger systems of equations ( i.e 5x5 ) so it is good to learn them.

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Yeah. But it is very helpfull later on when you start dealing with larger systems of equations ( i.e 5x5 ) so it is good to learn them.

Well, for IB at least that would be a paper 2 question since 5x5 matrices are not to be solved by hand. In other words, you can use the PolySmlt program on your calculator to solve it for you . But yeah I see what you mean though...

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I also find elimination personally much easier but substitution goes rather quick when one of the equations is already in single form (i.e y = and not 3y=).

As for using matrices, IF you have a calculator then it is much much easier and quick. However, SL does require you to know how to solve a 2x2 matrix equation using matrices without the calculator (came up on the specimen paper) so it's good to practise on that as well :/

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