How to find $x$ in $ax =b$ and $bx = c$, when $a$ and $c$ are fixed?

Question

In real terms: for $a = 27$ and $c = 1024$, I would like to find the numbers $x$ such that $$ 27x = b \text{ and } bx = 1024 \;, $$ like $3 \cdot 4=12 \cdot 4=48$. So, $12$ is $4$ times $3$ and $1/4$ of $48$. How do I find $4$ and $12$?

Answer 1

Since $ax=b$, then $bx=(ax)\cdot x$, so you get quadratic equation $ax^2=c$, or $ax^2-c=0$. Solve it and you will find your solutions.

Answer 2

if i'm right understand we get $$x=\frac{b}{27}$$ and $$\frac{b^2}{27}=1024$$ is that what you meant?

Answer 3

From $\color{blue}{27x} = \color{blue}{b}$ and $\color{blue}{b}x = 1024$ it follows that $$ \color{blue}{(27x)}x = 1024 \;. $$ If you now divide both sides by $27$, you will see that this expression is equivalent to $$ x^2 = \dfrac{1024}{27} \;. $$ Now you take the positive and the negative root, and you will obtain the two solutions $$ x_1 = +\sqrt{\dfrac{1024}{27}} = +\dfrac{32}{\sqrt{27}} \quad\text{and}\quad x_2 = -\sqrt{\dfrac{1024}{27}} = -\dfrac{32}{\sqrt{27}} \;. $$