If a, b, c are real numbers and b different from 0, find the relationships between a, b, c and the numbers q, r and s for which the linear system. post results. ax+by=c

qx+ry=s

(a) has no solutions.

(b) has infinitely many solutions.

(c) has exactly one solution.

Answer and Step-by-step explanation:

Given

ax + by = c

qx + ry = s

(a) the equation has no solutions if a/q = b/r ≠ c/s, when this happens, we say the system of equations has no solution. For example

x + y = 3

x + y = 5

Subtracting first equation from the second we have:

0 = 2 which is impossible.

(b) the equations have infinite solutions if a/q = b/r = c/s, for example

x + y = 2

x + y = 2

Subtracting the first equation from the second we have

0 = 0, since this is always true, then it has infinite solutions.

(c) the equations have unique solutions if a/q ≠ b/r, for example

x + y = 2

x – y = 1

Adding the first and second equation we have

2x = 3, we can get x from here and definitely y, so we have just one solution.

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

i need a diagram

step-by-step explanation:

no