Question

Solve the following system of equations.
2a – 3b + c = 10
2a – 2b – 2c = 2
a + 3b + 2c = –1

Answer 1

use determinant to solve above equation

Answer 2

idk wat that means

Answer 3

there are different methods that all come to the same results; since you have not defined a method, the determinant is suitable

Answer 4

the determinant is kidna useless you can tell if it's not factorable once you have the equation narrowed to 2 variables.

Answer 5

just use substitution :D!

Answer 6

ugh, substitution on these takes forever!!

Answer 7

id eliminate :)

Answer 8

how do u eliminatte

Answer 9

you scale the eqauations so that you can combine them and eliminate one of more of the variables

Answer 10

R1, 2a – 3b + c = 10
R2, 2a – 2b – 2c = 2
R3, a + 3b + 2c = –1

R3(-2), R2(-1)

R1, 2a – 3b + c = 10
R2, -2a + 2b + 2c =-2
R3, -2a -6b -4c = 2

R1+R2, and R1+R3


R1, 2a – 3b + c = 10
R2, 0 - b + 3c = 8
R3, 0 - 9b - 3c = 12

R2(-9)


R1, 2a – 3b + c = 10
R2, 0 +9b -27c = -72
R3, 0 - 9b - 3c = 12

R2 + R3


R1, 2a – 3b + c = 10
R2, 0 +9b -27c = -72
R3, 0 + 0 - 30c =-60

Answer 11

-30c = -60
c = 2

Answer 12

R1, 2a – 3b + (2) = 10
R2, 0 +9b -27(2) = -72

R1, 2a – 3b = 10-2
R2, 0 +9b = -72+54

b = -18/9 = -2


R1, 2a – 3(-2) = 8
R1, 2a + 6 = 8

a = (8-6)/2 = 1

Answer 13

2a – 3b + c = 10
1 -2 2
-------------
2 + 6 + 2 = 10


2a – 2b – 2c = 2
1 -2 2
--------------
2 + 4 - 4 = 2


a + 3b + 2c = –1
1 -2 2
------------------
1 -6 + 4 = -1

Answer 14

wow that is a lot of work
i do not like these so much do you?

Answer 15

i find them tedious

Answer 16

it takes longer to show the work than it does to actually get the answer :)

Answer 17

thankyou for ur help!

Answer 18

i hope it was helpful :)