Question

How can I solve using elimination and back substitution?

Answer 1

can some of u just help me with my stereo one

Answer 2

{4x-2y-4z=32
x-y+4z=-13
6x-4y+2z=14}

Answer 3

I already know the first step is 4x-2y-4z=32
x-y-4z=-13

Answer 4

5x-4y=19

Answer 5

I become lost after this step

Answer 6

Hello? Are u able youassist?

Answer 7

solve one equation for some variable, then plug in into one of the other equations, then solve for one of the two variables. You then have one of your variable. Though I think it is easier to use Gaussian elimination.

Answer 8

Gaussian elimination? Can you show me how that works?

Answer 9

write the coefficients of each variable for each equation into a matrix like pattern and add or subtract rows from one another until you have the pattern:

100=a
010=b
001=c

or

001=a
010=b
100=c

Answer 10

Not really seeing how that will help me get the correct answer

Answer 11

Its better because there is no substituting. You add and subtract rows until you end up with diagonal ones. If the one in the 'a' equation is in the x column, that is your x and so on.

Answer 12

I can help you. When you substitute, you first want to solve for a single variable. So take your equation 4x-2y-4z=32. You can solve for x by adding 2y and 4z to the other side. It is then 4x=2y+4z+32. Then divide by 4 making it x=1/2y+z+8

Answer 13

Now, you can plug that x into another equation. x-y-4z=-13 for instance. When you plug in what you've found for x, you'll only end up with z and y values in it. So (1/2y+z+8)-y-4z=-13

Answer 14

I just used the parentheses to show what the equation was for x. Is this making sense?