# Dependent Simultaneous Equations

My maths TB says:

For the simultaneous equations, $$a_1x + b_1y + c_1 = 0$$ and $$a_2x + b_2y + c_2 = 0$$ if $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$$ then the pair of equations are said to be consistent and dependent, and when plotted produce coincident lines.

I understand that these are consistent in the sense that they have some solution(though infinite).

But why are these called dependent?

An equation $$R(x_1,x_2, ... , x_n)=0$$ is called dependent on an equation $$S(x_1, x_2, ..., x_n)=0$$ if the equation $$R(x_1,x_2, ... , x_n)=0$$ can be derived algebraically (by using a finite number of the operations of addition, subtraction, multiplication, division, and exponentiation with constant rational exponents) from the equation $$S(x_1, x_2, ..., x_n)=0$$. If two equations are dependent on each other, we call each of them dependent.

According to the Merriam-Webster dictionary, dependent means

determined or conditioned by another.

So, calling such equations dependent seems reasonable because knowing each one can determine the other.

For example, let $$R(x,y)=a_1x+b_1y+c_1=0$$ and $$S(x,y)=a_2x+b_2y+c_2=0$$. Suppose that $$k=\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$ is a nonzero constant. Then we can derive $$R(x,y)=0$$ from $$S(x,y)=0$$ by multiplying both sides of $$S(x,y)=0$$ by $$k$$ as follows.$$k(a_2x+b_2y+c_2)=k(0) \quad \Rightarrow \quad a_1x+b_1y+c_1=0.$$Similarly, we can derive $$S(x,y)=0$$ from $$R(x,y)=0$$ by dividing both sides of $$S(x,y)=0$$ by $$k$$ as follows.$$\frac{a_1x+b_1y+c_1}{k}=\frac{0}{k} \quad \Rightarrow \quad a_2x+b_2y+c_2=0.$$Thus, the equations $$R(x,y)=0$$ and $$S(x,y)=0$$ are dependent.

$$\begin{bmatrix}a_1 & b_1 \\ a_2 & b_2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}c_1 \\ c_2\end{bmatrix}$$
Now, we say that the above system has infinite solutions if the two vectors $$(a_1, b_1)$$ and $$(a_2, b_2)$$ are linearly dependent (i.e, there exists some $$\lambda$$ such that $$(a_2,b_2) = \lambda(a_1, b_1)$$)