Question

If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and second equations, then the second and third equations.

3x – 5y + 2z = 7
4x + 9y – z = 2
x – 3y + 3z = 7


11x + 13y = 11
13x + 24y = 13

11x + 13y = 11
3x – 5y = 7

x – 3y = 7
13x + 24y = 13

x – 3y = 7
3x – 5y = 7

Answer 1

need help ASAP :\

Answer 2

eliminating from the 1st and second gives 11x+13y=11
but i cant get the second part right

Answer 3

In the second equation you have a -z and in the third equation you have a 3z. To get rid of the z, you need to make the second equation's z coefficient to be the opposite of the third equation's. What could you multiply the second equation by to get a -3 in front of the z?

Then add the second and third equation.

Answer 4

the second equation becomes 12x+27y-3z=6
subtracted: 13x+36y=8

Answer 5

am i rite?

Answer 6

oh wait...lemme try that again

Answer 7

subtracted: 13x+24y=-1

Answer 8

eliminate the x z in the second equation , you can find out the value for z in terms of x and y now substitute this value in 1st and third equation wyou will get 1) ans corerect

Answer 9

13x+24y=13...final

Answer 10

yea??