Applications of simultaneous equations

Question

Find the value of m for which the pair of simultaneous equations 3x + my = 5 and (m + 2)x + 5y = m have:

a) infinitely many solutions

b) no solutions.

Answer 1

a) m=-5 has infinite solutions

b) m=+3 has no solution

a) they have infinitely many solutions when they are the same equation, graphically, the same line

3x + my = 5

(m+2)x + 5y = m

the coefficients of the x and terms and the constant terms have the same ratio

5/m = 3/(m+2) = m/5

1st & last ratios

5/m = m/5

m^2 = 25,

m = + or - sqr25

m = -5 (or + 5, but ignore the positive square root)

(m+2)x +5y = m

(-5+2)x +5y =-5

-3x +5y = -5

3x +my = 5

3x +(-5)y = 5

3x -5y = 5 multiply by -1 to get -3x+5y =-5, which is the same as the other equation. Graphically, they are the same line. Algebraically, they are linearly dependent.

b) they have no solution if the lines have the same slope, if they are parallel, but not the same line

2nd & last ratios

3/(m+2) = m/5

m^2 +2m = 15

m^2 +2m -15 = 0

(m+5)(m-3) = 0

m =3 gives a parallel line, m=-5 gives the same line

m=3 gives a parallel line that never intersects, so there is no solution

(m+2)x + 5y = m

(3+2)x +5y = 3

5x +5y = 3

divide by 5 to get x+y =3/5

3x +my = 5

3x+3y =5

divide by 3 to get x+y =5/3

x+y=5/3 and x+y=3/5 are two parallel lines, both with slope =-1, but with different y intercepts

in slope intercept form they are y=(-1)x + 5/3 and y=(-1)x +3/5

both downward sloping straight lines that never intersect. With no intersection points, there is no solution to the two equations. Any (x,y) point that satisfies one equation will never satisfy the other equation.