Solving 'simple' system of linear equations

Question

I want to solve the following system of equations where $a \in \mathbb{R}$ is arbitrary

$$ax+12y=(a+16)x$$ $$12x+(a+7)y = (a+16)y$$

I know that a solution is $x = a$ and $y = \frac{4}{3}a$ but how do I derive this?

Answer 1

By distributing and moving over to the left side of the equal sign, the first equation is equivalent to

$$-16x+12y=0 \iff 4x-3y=0$$

The second is equivalent to

$$12x-9y=0\iff 4x-3y=0$$

So having the second equation introduces no new information. When this happens the system is dependent, i.e. possesses infinitely many solutions.

You can choose to let $x=t$ be an arbitrary value, and by substituting this into either equation, you find $y=\dfrac{4}{3}t$.

The solution you gave: $x=a, y=\dfrac{4}{3}a$ is a particular solution to this system, if we let $t=a$, but it's just one of an infinite number. For example, $(x,y)=(3,4)$ is a solution, as is $(0,0)$ and $(6a^3, 8a^3)$.

Answer 2

HINT: The first equation implies $$ ax + 12 y = ax +16 x \quad \Rightarrow \quad 12 y = 16x \quad \Rightarrow \quad y = \frac{4}{3} x. $$ Plug this equation into the second equation and solve for $x$.