simultaneous equations

Question

mx+3y=2

4x+(4+m)y=2m

a) find the values of m for which these are:

i) no solution

ii) infinitely many solutions

iii) unique solution

b) solve the equations in terms of m, for suitable values of m.

Answer 1

mx + 3y = 2

4x + (4+m)y = 2m

(4+m)/3 = 2m/2 = m

3m= 4+m

2m = 4

m = 2

m/4 = 3/(4+m) = 2/2m = 1/m ratios of the coefficients are constant

m^2 -= 4

m = 2

mx+3y = 2, 2x+3y =2

4x +(4+m)y = 2m, 4x+6y =4, divide by 2 to get 2x+3y =2

m=2 makes the 2 equations the same, with infinite solutions

m=0 gives a unique solution for (x,y) = (-2/3, 2/3)

4x +(4+m)y = 2m

4x +4y = 0

y =-x

mx + 3y =2

3y =2

y=2/3

x=-2/3

that's equivalent to a downward slopeing line through the origin with slope =-1 and a horizontal line y= 2/3

to get no solution, you need to make the two lines parallel

mx + 3y = 2

3y = -mx +2

y = -mx/3 + 2/3

4x + (4+m)y = 2m

(4+m)y = -4x +2m

y = -4x/(4+m) + 2m/(4+m)

to be parallel the coefficients of the x terms must be equal

-m/3 = -4/(4+m)

-12 = -4m -m^2

m^2 +4m -12 = 0

(m+6)(m-2) = 0

m = 2 or -6

2 makes the two lines the same with infinite solutions

-6 makes the lines parallel with no solution