Question

HELP PLEASE!! MEDALS WILL BE GIVEN!!

Solve the system of equations:
9a+3b+c=2
a+b+c=4
6a+b=0

Answer 1

Pick one equation and solve for one variable. It doesn't matter which one. Show me your work and I will give you the next step.

Answer 2

b=-6a

Answer 3

Great. Now substitute -6a for b in one of the other equations.

Answer 4

9a + 3(-6a) +c =2

Answer 5

so that equals to -9a+c=2

Answer 6

Now substitute the same -6a for b in the other 3 variable equation. That should leave you with 2 equations and 2 variables.

Answer 7

a + -6a +c =4
-5a +c =4

Answer 8

so then c= 4 +5a

Answer 9

Don't solve for c on this one. Your two equations are:
-9a+c=2
-5a+c=4
Can you solve a systems of equations with two variables? I would suggest subtraction....

Answer 10

alright thank you :) i got it from here :)

Answer 11

Great. If you need an additional resource, www.purplemath.com is a great resource for high school math.