Solve for m and n:

2m-n=3

mē - mn+nē =3 :mad:

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- Aug 8th 2005, 10:19 AMflyStuck
Solve for m and n:

2m-n=3

mē - mn+nē =3 :mad: - Aug 8th 2005, 10:52 AMticbol
Here is one way, by substitution.

2m -n = 3 ------(1)

mē -mn +nē = 3 ---------(2)

From (1),

2m -3 = n, or, n = 2m -3

Substitute that into (2)

m^2 -m(2m -3) +(2m -3)^2 = 3

m^2 -(2m^2 -3m) +(4m^2 -12m +9) = 3

m^2 -2m^2 +3m +4m^2 -12m +9 -3 = 0

Collect like terms,

(m^2 -2m^2 +4m^2) +(3m -12m) +(9 -3) = 0

3m^2 -9m +6 = 0

Get its simplest form, divide both sides by 3,

m^2 -3m +2 = 0

Factor that,

(m -2)(m-1) = 0

m-2 = 0

So, m= 2

m -1 = 0

So, m = 1

When m=2,

n = 2m -3 = 2*2 -3 = 1

When m=1,

n = 2m -3 = 2*1 -3 = -1

Therefore, there are two sets of answer:

m=2, n=1 ----------answer.

m = 1, n = -1 ----------answer.