SOLUTION: how can i solve this system of equations using the addition method when one or more of the terms is missing? 4a+9b=8 8a+6c=-1 6b+6c=-1

Question 755962: how can i solve this system of equations using the addition method when one or more of the terms is missing?
4a+9b=8
8a+6c=-1
6b+6c=-1
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your original equations are:

4a+9b=8 (equation 1)
8a+6c=-1 (equation 2)
6b+6c=-1 (equation 3)

fill in the missing variables with 0 times that variable to get:

4a + 9b + 0c = 8 (equation 1)
8a + 0b + 6c = 1 (equation 2)
0a + 6b + 6c = 1 (equation 3)

now you have 3 equations in 3 unknowns which can be solved.

work on the first 2 equations and eliminate the a variable by multiplying the first equation by 2 and then subtracting the second equation from it.

after you multiply equation 1 by 2, it becomes equation 4.

you get:

8a + 18b + 0c = 16 (equation 4)
minus:
8a + 0b + 6c = 1 (equation 2)
equals:
0a + 18b - 6c = 15 (equation 5)

the result of the subtraction becomes equation 5.

since the a variable is already eliminated from equation 3, you don't have anything to do except take the remaining 2 equations (equation 5 and equation 3) and solve them as 2 equations in 2 unknowns.

you have:

18b - 6c = 15 (equation 5)
6b + 6c = 1 (equation 3)

since you have + 6c in equation 3 and - 6c in equation 5, all you have to do is add these equations together and you will have eliminated the c variable which will allow you to solve for b.

you get:

18b - 6c = 15 (equation 5)
plus:
6b + 6c = 1 (equation 3)
equals:
24b = 16

solve for b to get:

b = 2/3

you now have the value of b which you can now use to solve for c in equations 5 or 3.

either one will do, and you should get the same answer regardless of which one you choose.

i'll do both to show you.

equations 5 and 3 are:

18b - 6c = 15 (equation 5)
6b + 6c = 1 (equation 3)

replace b with 2/3 in each of these questions and solve for c to get:

c = -(1/2) in equation 5.
c = -(1/2) in equation 3.

now you have:

b = 2/3
c = -(1/2)

go back to your original 3 equations and use the values of b and c to solve for a in any one of those equations that has a in it.

you only need to solve for a in one of those equation and then you can confirm by using the values of a,b,c in all 3 equations. if the equations are true, then you did good.

equation 1 states that 4a + 9b = 8
if b = 2/3, then this equation becomes:
4a + 9(2/3) = 8 which becomes:
4a + 6 = 8 which becomes:
4a = 2 which becomes:
a = (1/2)

you now have:

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

plug these values in all 3 original equations and they should all be true.
if so, you're done.
if not, you messed up somewhere and have to go back and see where you went wrong.

the original 3 equations are:

4a + 9b = 8 (equation 1)
8a + 6c = 1 (equation 2)
6b + 6c = 1 (equation 3)

replacing a with 1/2 and b with 2/3 and c with -(1/2), get you:

4(1/2) + 9(2/3) = 8
8(1/2) + 6(-(1/2)) = 1
6(2/3) + 6(-(1/2)) = 1

solving each of these equations gets:

8=8 (true)
1=1 (true)
1-1 (true)

you did good.

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