Today we learned how to solve systems of equations with 3 variables, however my teacher didn't go over how to do a problem where not every single equation in the system has the 3 variables. Can someone point me to a place that explains this?
the kind of problem im talking about is like this
solve the system of equations:
6a-2b=18
3b+5c=-34
a+6c=-28
notice that each of the equations is missing a different variable.
If we can eliminate the variable "a" from the 1st and 3rd, we have a new equation with b's and c's like the 2nd equation
so. from the 3rd a = -6c - 28
sub that into the 1st
6(-6c - 28) - 2b = 18
-36c - 168 - 2b = 18
36c + 2b = -186
b + 18c = -93
let's multiply that by 3
3b + 54c = -279
subtract the 2nd
49c = -245
c = -5
then a = -6(-5) - 28 = 2
and
3b + 5(-5) = -34
3b = -9
b = -3
a = 2 , b = -3 , c = -5
When solving systems of equations, it can sometimes happen that not every equation in the system has all the variables present. In this case, you can still solve the system by using different methods, such as substitution or elimination.
Let's take a look at the system of equations you provided:
6a - 2b = 18 ...........(1)
3b + 5c = -34 ...........(2)
a + 6c = -28 ...........(3)
To solve this system, we will use the elimination method. However, before proceeding, we need to look for a way to eliminate one variable. Let's start by looking for a pair of equations that allow us to eliminate either 'a' or 'b'.
By observing equations (1) and (3), we can see that we can eliminate 'a'. To do that, we will multiply equation (3) by 6:
6(a + 6c) = 6(-28)
6a + 36c = -168 ...........(4)
Now, let's consider this new equation (4) alongside equation (1):
6a + 36c = -168 ...........(4)
6a - 2b = 18 ...........(1)
By subtracting equation (1) from equation (4), we can eliminate 'a':
(6a + 36c) - (6a - 2b) = (-168) - (18)
2b - 36c = -186
Therefore, we have a new equation:
2b - 36c = -186 ...........(5)
Now we have two equations:
2b - 36c = -186 ...........(5)
3b + 5c = -34 ...........(2)
We can solve this system of equations using elimination or substitution as you have previously learned. In this case, let's use elimination again.
To eliminate 'b', we'll start by multiplying equation (2) by 2:
2(3b + 5c) = 2(-34)
6b + 10c = -68 ...........(6)
Now, let's consider this new equation (6) alongside equation (5):
2b - 36c = -186 ...........(5)
6b + 10c = -68 ...........(6)
By multiplying equation (5) by 3 and equation (6) by 1, we can create equal coefficients for 'b':
6(2b - 36c) = 6(-186)
3(6b + 10c) = 3(-68)
This simplifies to:
12b - 216c = -1116 ...........(7)
18b + 30c = -204 ...........(8)
Now we have two equations:
12b - 216c = -1116 ...........(7)
18b + 30c = -204 ...........(8)
We can solve this system of equations using elimination or substitution, but you might notice that we could simplify each equation by dividing by their respective coefficients. This can make the process easier and avoid large numbers:
(12b - 216c = -1116) / 12 becomes (b - 18c = -93) ...........(9)
(18b + 30c = -204) / 6 becomes (3b + 5c = -34) ...........(10)
Now we have simplified equations (9) and (10) with the same variable 'b'. We can solve this new system of equations using either elimination or substitution, which you have already learned.
By solving equations (9) and (10) using your preferred method, you will find the values of 'b' and 'c'. Then, substituting these values back into equations (1) or (3), you can solve for 'a'.
Remember, the key is to simplify the system by eliminating variables or creating equal coefficients before solving the system using the methods you already know.