# Thread: Systems of Linear Equations

1. ## Systems of Linear Equations

Hello all,

i am studding linear Algrabra from my finals in about 5 days :-)

I have 2 systems that i dont know exactly how to solve them. Would any1 like to help me solve them ?

I'd like step by step solution if its possible

the 1st system is this one:

dx+y+z=a
x+dy+z=b
x+y+dz=c

where a,b,c Real Numbers

The Second is this one:

ax+by+z=1
x+aby+z=b
x+by+az=1
Where a,b Real Numbers

Please DONT solve them both, just one of them,
so i ll try the second on my own and then i ll post my
solution, so you can correct my mistakes

Thanks alot in advance for your help people , its really important :-)

2. ## Re: Systems of Linear Equations

Originally Posted by primeimplicant
Hello all,

i am studding linear Algrabra from my finals in about 5 days :-)

I have 2 systems that i dont know exactly how to solve them. Would any1 like to help me solve them ?

I'd like step by step solution if its possible

the 1st system is this one:

dx+y+z=a
x+dy+z=b
x+y+dz=c

where a,b,c Real Numbers

The Second is this one:

ax+by+z=1
x+aby+z=b
x+by+az=1
Where a,b Real Numbers

Please DONT solve them both, just one of them,
so i ll try the second on my own and then i ll post my
solution, so you can correct my mistakes

Thanks alot in advance for your help people , its really important :-)
dx+y+z=a
x+dy+z=b
x+y+dz=c

z=a-y-dy

Put this z into
x+dy+z=b
x+y+dz=c

And solve system of 2 equations with 2 variables.

3. ## Re: Systems of Linear Equations

Originally Posted by Also sprach Zarathustra
dx+y+z=a
x+dy+z=b
x+y+dz=c

z=a-y-dy

Put this z into
x+dy+z=b
x+y+dz=c

And solve system of 2 equations with 2 variables.
hmmm you actually mean z=a-y-dx Right ?

Isnt it quite easy mode that way ?

Isnt there any more complex way to solve this ?

Would you like to give me some solution, the way you would do this? =/

Thanks again for your help

4. ## Re: Systems of Linear Equations

Originally Posted by primeimplicant
hmmm you actually mean z=a-y-dx Right ?

Isnt it quite easy mode that way ?

Isnt there any more complex way to solve this ?

Would you like to give me some solution, the way you would do this? =/

Thanks again for your help
Yes!

Yes!

Yes! But way to complex things?

I gave you.

You also can solved it with Cramer's rule - Wikipedia, the free encyclopedia.

Good-luck!

5. ## Re: Systems of Linear Equations

Hello again, i solved the first expample in Wikipedia correctly, but i have big trouble in sovling this one, that is more abstract (i mean without numbers). This d, is really confusing me, and theres no inidication if its a real number or something...

here is what i have done so far

ill try now the second equation

edit:

Alright i stuck on second as well -.-

this is what ive done so far, i might be tired and cant see the obvious... unlsess is something else tha requires more knowledge.... tips ?

6. ## Re: Systems of Linear Equations

ax+by+z=1
x+aby+z=b
x+by+az=1

If you subtract the first equation from the second you eliminate z: (1- a)x+ (ab-b)y= b-1.

If you subtract a times the first equation from the third equation you also eliminate z: (1-a^2)x+ (b- ab)y= 1- a

Note that b-ab= -(ab-b) so if you add those two equations, you eliminate y: (1- a^2+ 1- a)x= b-1+ 1- a or
(a^2+ a- 2)x= a- b.

Can you finish that?

7. ## Re: Systems of Linear Equations

Originally Posted by HallsofIvy
ax+by+z=1
x+aby+z=b
x+by+az=1

If you subtract the first equation from the second you eliminate z: (1- a)x+ (ab-b)y= b-1.

If you subtract a times the first equation from the third equation you also eliminate z: (1-a^2)x+ (b- ab)y= 1- a

Note that b-ab= -(ab-b) so if you add those two equations, you eliminate y: (1- a^2+ 1- a)x= b-1+ 1- a or
(a^2+ a- 2)x= a- b.

Can you finish that?
thanks for your help, ill try finish this