Algebra II: missing variables

[Tay]

Can you help with this... i have to solve for x,y,z... i'm missing variables and my book says it's supposed to be easier with the variables missing??

3x + 4y = 19
2y + 3z = 8
4x -5z = 7

 

[Mrspi]

Can you help with this... i have to solve for x,y,z... i'm missing variables and my book says it's supposed to be easier with the variables missing??

3x + 4y = 19
2y + 3z = 8
4x -5z = 7

I don't see that there are any "missing variables."

3x + 4y + 0z = 19
0x + 2y + 3z = 8
4x + 0y - 5z = 7

If you can solve a system like this when all three variables are present, you should be able to solve this system, too.

Hint....

The first equation is one involving just x and y.....can you combine the second and third equations to get another that involves just x and y? (multiply the second equation by 5, and the third equation by 3....add the results)

5*0x + 5*2y + 5*3z = 5*8
0x + 10y + 15z = 40

3*4x + 3*0y - 3*5z = 3*7
12x + 0y - 15z = 21

Or....
10y + 15z = 40
12x - 15z = 21

Add those together...you will get

12x + 10y = 61

Now...use THAT, and the first equation in your system, which involved only x and y:

3x + 4y = 19
12x + 10y = 61

Ok...you take it from here.

If you are still having trouble, please repost, showing what you've done.