Solving systems of lin. eqns: x+5y+14z=1,5x+26y+73z=5,-2y-6z

[Angela123]

Solve the following system of linear equations:

x+ 5y+ 14z=1
5x+26y+ 73z=5
-2y -6z=0
4x+19y+53z=4

I did this:
-5(x+5y+14z=1)-->-5x-25y-70z=-5--> y+3z=0
5x+26y+73z=1 --> 5x+26y+73z=5
Then, I did:
y+3z=0-->2(y+3z=0)-->2y+6z=0
-2y-6z=0-->-2y-6z=0 -->-2y-6z=0
Since it all cancels out, I thought y=0, z=0, but that's not right. Does anyone know what you do when it all cancels out like that?

 

[Denis]

Re: linear equations

x + 5y + 14z = 1 [1]
5x + 26y + 73z = 5 [2]
-2y - 6z = 0 [3]
4x + 19y + 53z = 4 [4]

Divide [3] by -2 : y + 3z = 0 ********

5x + 26y + 73z = 5 [2]
5x + 25y + 70z = 5 [1] * 5
Subtract: y + 3z = 0 *******

4x + 20y + 56z = 4 [1] * 4
4x + 19y + 53z = 4 [4]
Subtract: y + 3z = 0 *******

******* so the equations you're given are "useless"; there's an infinity of solutions.

 

[Angela123]

Re: linear equations

Yea, I don't understand it either. I do my homework online, and I tried just putting a zero in for the ones with infinitely many solutions but it says it's wrong. So was I going about solving it the right way before?

 

[mmm4444bot]

Re: linear equations

Were you able to get x - z = 1?

The system reduces to two equations.

y + 3z = 0

x - z = 1

We can express both x and y in terms of z by solving for x and y.

 

[Denis]

Re: linear equations

....and I tried just putting a zero in for the ones with infinitely many solutions....

Don't know what you mean with that, Angela.
There is an infinite number of solutions, period.
In this style (x,y,z):
.....
-4 15 -5
-3 12 -4
-2 9 -3
-1 6 -2
0 3 -1
1 0 0
2 -3 1
3 -6 2
4 -9 3
.....
You can go on indefinitely going North or South :shock:

 

[Angela123]

Re: linear equations

Thanks, I finally figured out that it wanted me to enter x=z+1 instead of an actual number to show infinite solutions. I appreciate you taking the time to help!