# Show a problem with a system of dependent equations.

### 2 Answers | Add Yours

The simplest system of dependent linear equations of two variables x and y is: ax+by = c and k(ax+by) = kc.

The equations of this type could be derived from one another by a suitable multiplication or division by a constant.

Since both equations represent same equation, the coordinates of all the points on the graph are the solutions. So there are infinitely many solutions.

The condition for dependent equations a1x+b1x+c1 = 0 and a2 x+b2x+c2 = 0 is:

a2/a1 = b2/b2=c2/c1 = l a finite number.

Example: x+2y = 3 and 3x+6y = 3.

Both equations could be written like :

x+2y-3 = 0 and 3x+6y-9 = 0.

3/1= 6/2 = -9/-3 = 3.

Therefore they are dependent system of equations.

The solutions are many:

Some solutions are:

(x,y) = (1,1), (2, -1/2), (5 , -1), ( 7, -2) etc.

Dependent equations describe the same line, though they are different. The system formed by dependent equations has an infinite number of solutions.

For instance:

3x + 4y = 2

6x + 8y = 4

We notice that we get the second equation multiplying by 2 the first equation.

So, both equations are equivalent and all points located on a line, they are located on the other line also.

**The system has an infinite number of solutions.**