Question

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

Answer 1

do u know how to use matrix method?

Answer 2

first one a=-c/2+3b/2+5

Answer 3

no :/

Answer 4

second one a=b+c+1

Answer 5

third one b=-2c/3-1/3

Answer 6

did that help

Answer 7

lauren are u still there

Answer 8

or u can use this way
2a – 3b + c = 10 R1
2a – 2b – 2c = 2 R2
a + 3b + 2c = –1 R3
keep R1
R2' = R1-R2 (illumination method)
= -b + 3c = 8
R3' = R1-2R3 (illumination method)
= -9b -3c = 12
now the system changed to
2a-3b+c = 10 R1
-b+3c = 8 R2'
-9b-3c = 12 R3'
one more time, now we keep R1 and R2'
R3''= R2'+R3'
= -10b = 20
finally the system changed to
2a-3b+c = 10 R1
-b+3c = 8 R2'
-10b=20 R3''
now u can get b from R3'', then use the back-substitution to find a and c
The answer are: b=-2, c=2 , a= 1

Answer 9

if u know how to use matrix method, it would be easier than this way