Linear Equations: Solving for 3 unknowns given 3 equations

[kaydis]

Homework Statement

I need to work out the values of a, b, and c

Relevant Equations

a+b=0
-2a-2c = -1
-a-3b+c= 1

a+b=0

so..
a= -b

-2a-2c = -1 = -2(-b)-2c = -1 = 2b-2c=-1

-a-3b+c= 1 = -(-b)-3b+c= 1 = b-3b+c=1 = -2b+c=1

i think this is right but i don't know where to go from here

 

[berkeman]

a= -b

-2a-2c = -1 = -2(-b)-2c = -1 = 2b-2c=-1

-a-3b+c= 1 = -(-b)-3b+c= 1 = b-3b+c=1 = -2b+c=1

Good work! You've been able to reduce the system of equations from 3 equations and 3 unknowns to 2 equations with 2 unknowns. What can you do to combine those last two equations to eliminate the "b" terms?

 

[berkeman]

BTW, remember that after you have solved for a, b and c, it's a good idea to plug those back into the original 3 equations that you started with to be sure they are satisfied by the solutions you found.

 

[kaydis]

Good work! You've been able to reduce the system of equations from 3 equations and 3 unknowns to 2 equations with 2 unknowns. What can you do to combine those last two equations to eliminate the "b" terms?

could i add them?

2b + -2b = canceled out
and
-2c + c = -c
and
-1 + 1 = 0
so
-c=0

 

[berkeman]

Yep, good job. Now solve for b, then solve for a, and check that your values work in the original equations. Can you show us that?

 

[kaydis]

I think i did it!
-c=0

-2a-2c=-1 = -2a-2(0)=-1 = -2a=-1 = -a=-1/2 or -0.5

SO a=0.5

-a-3b+0=1 = -a-3b=1 = -0.5-3b=1 = -3b=1.5 = -b=1/2 or 0.5

SO b=-0.5

when i plug them back in i get the following:

a+b=0 --------> 0.5+-0.5=0
-2a-2c=-1 --------> -2(0.5)-2(0)=-1
-a-3b+c=1 --------> -0.5-3(-0.5)+0=1

 

[berkeman]

Looks good!

 

[kaydis]

awesome, thank you so much for all the help!