tough question in math ..system of equatiions?

Question

conssider the system of equations
a + b = 2m^2
b+c=6m
a+c =2

Determine all real values of m for whcih a<=b<=c

Answer 1

In order to find the range of values for m that apply, you first have to get each variable in terms of m....

Multiply the 2nd equation by (-1) and add it to the 3rd one:

a - b = 2 - 6m

Now, add this simplified result to the 1st eqn:

2a = 2m^2 - 6m + 2

a = m^2 - 3m + 1


Now that we have a in terms of m, we can get b in terms of m as well.....

b = 2m^2 - a = 2m^2 - (m^2 - 3m + 1)

= m^2 + 3m - 1


Repeat the same process for variable c, using rearranged versions of either Equation 2 or 3......we'll use Equation 3 for the hell of it:

c = 2 - a = 2 - (m^2 - 3m +1)

= 1 + 3m - m^2


Now we can establish the acceptable range of values for m:

If a <= b, then.....

m^2 - 3m +1 <= m^2 + 3m - 1

2 <= 6m ---------------> m >= 1/3

We aren't done yet though, because b <= c as well.....

m^2 + 3m - 1 <= 1 + 3m - m^2

2m^2 <= 2

m^2 <= 1

m <= 1, -1***

***Note: m <= -1 has to be eliminated as an answer, because we previously established that m >= 1/3

These facts give us the acceptable range for m:

1/3 <= m <= 1