Question

solve the equations

Answer 1

Hmm maybe. Lets see!

Answer 2

to start it off, you have to solve for the equation of a, b, and c first. Ill do the first equation. One question. Are you you doing elimination/substitution in your class right now?

Answer 3

both

Answer 4

2A+3B-2C=0
-3B -3B
___________________________
2A-2C=-3B
+2C +2C
_______________________
2A= -3B+2C
\[\frac{ 2A }{ 2 }=\frac{ -3B+2C }{ 2 }=A=\frac{ -3 }{ 2}B+C\]

Answer 5

okay thats great. We'll be using a lot of substitution.

Answer 6

Do you understand what I did above?

Answer 7

okya

Answer 8

yes

Answer 9

you have 2 equations and 3 unknowns - so this cannot be solved.

generally if you have N unknowns, then you need N equations to get a unique solution.

Answer 10

uless you have been asked to find just the integer solutions?

Answer 11

*unless

Answer 12

i also think that cant be solve
what u r trying to do u have 3 variables ans just 2 equations

Answer 13

Okay thats the equation for A. We'll substitute that for A is the original equation. Next were gonna find B.

2(-3/2B+C)+3B-2C=0

Answer 14

-3B+2C=0
-2C -2C
--------------
-3B=-2C
\[\frac{ -3B }{ -3}=B =\frac{ -2 }{ -3 }C=\frac{ 2 }{ 3 }C\]
SO B= \[\frac{ 2 }{ 3 }C\]

Answer 15

Now were gonna plug that into the original.
2A+3B-2C=0
2(-3/2B+C)+3(2/3C)-2C=0
-3B+2C+2C-2C=0
-3B+2C=0
+3B +3B
____________________
2C=3B
C=3/2B

Answer 16

I think thats all we can do here, we cant solve for variables.

Answer 17

i knw

Answer 18

but i wanna appreciate ur efforts that u have solved my sum though it can;t be solved as everyone said

Answer 19

:) Thanks! Have a good one!