Question

Solve the system by elimination;
4a - 2b + 3c = -10
5a + 2b - 2c = -28
-3a - 2b + c = 22

Answer 1

pic one variable of one equation and solve for it.

LIke pick c of bottom most equation and solve for it then plug it into middle equation

Answer 2

how do you eliminate the a and b though?

Answer 3

i end up with
4a - 2b + 3c = -10
5a + 2b - 2c = -28
-3a - 2b + c = 22

4a - 2b + 3c = -10
5a + 2b - 2c = -28
c = 22 + 3a +2b

Answer 4

take the first two equations and add them...this will cancel out the b's

Answer 5

https://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/fast-systems-of-equations/v/solving-systems-of-equations-by-elimination

Answer 6

wait i was doing substitution sorry @ilikequadraticsokay

Answer 7

the link i gave has a great example

Answer 8

4a - 2b + 3c = - 10
5a + 2b - 2c = - 28
-----------------
9a + c = - 38

now take the next two equations and eliminate b.
5a + 2b - 2c = - 28
-3a - 2b + c = 22
------------------
2a - c = - 6

now take the 2 equations you came up with...the ones you cancelled out the b's in.
9a + c = - 38
2a - c = - 6
-----------
11a = - 44
a = - 4

now sub -4 in for a in either of the above equations
9a + c = - 38
9(-4) + c = - 38
- 36 + c = - 38
c = - 38 + 36
c = - 2

now sub known values into any of the original equations
4a - 2b + 3c = -10
4(-4) - 2b + 3(-2) = - 10
-16 - 2b - 6 = - 10
-2b = - 10 + 22
-2b = 12
b = -6

lets check...
5a + 2b - 2c = - 28
5(-4) + 2(-6) - 2(-2) = - 28
-20 - 12 + 4 = - 28
- 32 + 4 = - 28
- 28 = - 28 (correct)

a = -4, b = -6, c = -2

Is there anything I have just done that you don't understand ? I will be happy to explain it to you.

Answer 9

You explained this perfectly, thank you!

Answer 10

your welcome :)