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

(eq1) 3x – 5y + 2z = 7
(eq2) 4x + 9y – z = 2
(eq3) x – 3y + 3z = 7


(eq1) 3x – 5y + 2z = 7
(eq2) 4x + 9y – z = 2
multiply whole eq2 by 2 gives eq2.1
(eq2.1) 8x + 18y -2z = 4
add eq2.1 to eq1

So:
(eq1) 3x – 5y + 2z = 7
(eq2.1) 8x + 18y -2z = 4
= 11x + 13y = 11 <----

(eq2) 4x + 9y – z = 2
(eq3) x – 3y + 3z = 7
next, multiply all of eq2 by 3, giving eq2.2
(eq2.2) 12x + 27y – 3z = 6

Then add eq2.2 and eq3 together
(eq2.2) 12x + 27y – 3z = 6
(eq3) x – 3y + 3z = 7
=13x + 24y = 13 <-----

which is the first pair.

Answer 2

Hope someone proofs this ... i think it is right though. :)

Answer 3

this is elimination using the eq2 for each one, and then eq2 is eliminated

Answer 4

thank you

Answer 5

when you do it in the future though, make sure you only multiply through using one equation, and then that is the equation removed