Question

If 12a+3b=1 and 7b–2a=0, what is the average of a and b?

Answer 1

Answer:

The mean/average of a and b is 1/20.

Step-by-step explanation:

Recall that the average of two numbers is equal to the sum of the numbers divided by 2.  Thus, we need to find (a+b)/2.

We'll first solve the system

12a+3b=1

7b–2a=0

I used matrices to solve this system:  a=7/90 and b = 1/45.

Another possible approach would be to use elimination by  addition and subtraction to solve this system.  Do 7/90 and 1/45 satisfy 7b-2a=0?  Yes.  Do 7/90 and 1/45 satisfy 12a+3b=1?  Yes.

Now we need to find (a+b)/2:  

7/90 + 1/45          7/90 + 2/90         9/90

-------------------- = -------------------- = ----------- = 1/20

      2                             2                      2

Answer 2

The average of 'a' and 'b' is evaluated to 0.05

What is a solution to a system of equations?

For a solution to be solution to a system, it must satisfy all the equations of that system, and as all points satisfying an equation are in their graphs, so solution to a system is the intersection of all its equation at single point(as we need common point, which is going to be intersection of course)(this can be one or many, or sometimes none)

For the given case, the system of equation we've got is:

$12a+3b=1\\7b-2a=0$

From the first equation, we get:

$12a+3b=1\\12a= 1-3b\\\\a=\dfrac{1-3b}{12}$

Putting this value of 'a' in second equation, we get:

$7b - 2a = 0\\7b - 2(\dfrac{1-3b}{12}) = 0\\\\7b - \dfrac{1-3b}{6} = 0\\\\42b - 1 + 3b = 0\\b = \dfrac{1}{45}$

Therefore, putting this value in expression we got for 'a', we get:

$a = \dfrac{1-3b}{12} = \dfrac{1-3(1/45)}{12} = \dfrac{14}{180} = \dfrac{7}{90}$

The average of two values is half of their sum, therefore, the average of 'a' and 'b' is evaluated as:

$A = \dfrac{a+b}{2} = \dfrac{7/90 + 1/45}{2} =\dfrac{9}{180} = \dfrac{1}{20} = 0.05$

Thus, the average of 'a' and 'b' is evaluated to 0.05

Learn more about solution of system of linear equations here:

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