simultNEOUS EQUAN

freeze4real

Consider the system of linear equations in X and Y
x + 2y = 6
2X + 4Y = 12

This system of equations has a unique solution.

i have solved this with simulataneous equation and i get no answer, am i right ?

Also how do isolve this using matrices?

thanks

TheBody

freeze4real wrote: »

Consider the system of linear equations in X and Y
x + 2y = 6
2X + 4Y = 12

This system of equations has a unique solution.

i have solved this with simulataneous equation and i get no answer, am i right ?

Also how do isolve this using matrices?

thanks

The system does NOT have a unique solution. It has an infinite number of solutions. You should hopefully notice that the second equation is just 2 times the first one. You can think about them as two lines lying on top of each other. As this means they intersect in an infinite number of places, the system has an infinite number of solutions.

With regards the solution set, as both equations essential represent the same line, you only really need to worry about one of the equations, say x+2y=6. If we re-arrange it we get x=6-2y. If we let y=t where t is a real number, then we can say that x=6-2t. So all the solutions are given by those formulae. For example, say when t=2, this gives us that x=6-2(2)=2 and y=2. Hopefully you can see that (2,2) satisfies your origional equations.

freeze4real

TheBody wrote: »

The system does NOT have a unique solution. It has an infinite number of solutions. You should hopefully notice that the second equation is just 2 times the first one. You can think about them as two lines lying on top of each other. As this means they intersect in an infinite number of places, the system has an infinite number of solutions.

With regards the solution set, as both equations essential represent the same line, you only really need to worry about one of the equations, say x+2y=6. If we re-arrange it we get x=6-2y. If we let y=t where t is a real number, then we can say that x=6-2t. So all the solutions are given by those formulae. For example, say when t=2, this gives us that x=6-2(2)=2 and y=2. Hopefully you can see that (2,2) satisfies your origional equations.

wow, simply wow that was a massive fail on my part, the way you did it can it be done through matrices ?

bnt

freeze4real wrote: »

wow, simply wow that was a massive fail on my part, the way you did it can it be done through matrices ?

If you try and do it using matrices, you get an invalid answer. You need to invert the first matrix, which involves finding the inverse of its determinant, but the determinant is zero - so it's a bust.

TheBody

freeze4real wrote: »

wow, simply wow that was a massive fail on my part, the way you did it can it be done through matrices ?

You can use Gauss-Jordon elimination if you want but it quickly boils down to what I did above. If you write down the augmented matrix and do row 2 - 2xrow 1, the second row will turn into a row of zeros. So you will just be left with row one which is just the first equation that I solved above.

freeze4real

TheBody wrote: »

You can use Gauss-Jordon elimination if you want but it quickly boils down to what I did above. If you write down the augmented matrix and do row 2 - 2xrow 1, the second row will turn into a row of zeros. So you will just be left with row one which is just the first equation that I solved above.

thanks,

my maths is poor and i want to do very well in my econometrics exam, so i was thining of solving this in multile ways.


TheBody you said that (2 ,2) satisfies the equation, and thats because you assumpted y = t = 2, what if y = t isnt 2 ?

TheBody

freeze4real wrote: »

thanks,

my maths is poor and i want to do very well in my econometrics exam, so i was thining of solving this in multile ways.


TheBody you said that (2 ,2) satisfies the equation, and thats because you assumpted y = t = 2, what if y = t isnt 2 ?

I didn't really assume that t=2. In what I did solving the problem I said that t could be ANY real number at all. Try it yourself.....pick any value you like for t and you will get a pair of values, (x,y) that will satisfy the equations. I just choose t=2 for an example. This is why there are an infinite number of solutions.

freeze4real

TheBody wrote: »

I didn't really assume that t=2. In what I did solving the problem I said that t could be ANY real number at all. Try it yourself.....pick any value you like for t and you will get a pair of values, (x,y) that will satisfy the equations. I just choose t=2 for an example. This is why there are an infinite number of solutions.

beautiful, there are infinite amount of solutions,

I tried x = 2y = 6

x= 6 - 2y

stating y=t=6

= 6 -2[6]
x = 6 - 12 = -6
x = -6

using this in the original equation,

x +2y = 6
-6 + 2(6) = 6
-6 + 12 = 6

thanks alot.

would it be possible to get the another answer by graphing it ?

TheBody

If you graph it you would just get two lines on top of each other. This illustrates how there are an infinite number of solutions as the lines intersect at an infinite number of points.

freeze4real

TheBody wrote: »

If you graph it you would just get two lines on top of each other. This illustrates how there are an infinite number of solutions as the lines intersect at an infinite number of points.

Yup, just did that now, and it looks very good.

Thanks a mill man

You're a legend.

TheBody

freeze4real wrote: »

Yup, just did that now, and it looks very good.

Thanks a mill man

You're a legend.

lol, your welcome!! Glad I could help.