Solving a system of three equations: $d = s\cdot 3, c = s\cdot 1.5, c = 2\cdot d$. Sorry if this is not the right place for this sort of question, but I am at a lost. My niece has some summer homework, and neither of us have a clue how to solve this question. Its been too long since I was in school, and my math skills have plummeted since.
Could someone please show me how to solve this, with step by step, or even a link to a similar question that I could follow. I need to find the value of $s$.
$$\begin{align}
d &= s\cdot 3\\
c &= s\cdot 1.5\\
c &= 2\cdot d
\end{align}$$
 A: Ah, well; If I gave you a system of two equations in two unknowns, $$ u=v, 3u=3v,$$ you would correctly answer that they could be anything, as long as they were the same.
If I switched to $$ x=y, x=5y   $$ you would be right to be suspicious. Indeed, we get $$ 5y=y,  $$ then subtract $y$ from both sides, $$ 4y=0,   $$ so $y=0$ and $x=0.$
My original reading of your system (i was not careful reading) was of the first type, $$  w = 3 u, v = 1.5 u, w = 2 v,     $$ which has infinitely many solutions, as long as the ratios are maintained, for example $(u=2,v=3,w=6),$ or
 $(u=22,v=33,w=66).$
However, as the others have pointed out, your system is of the inconsistent ratio type, the only answer is all equal to $0.$ If the third equation of your system were switched to $d=2c$ you would get infinitely many solutions. 
A: Since $c=2d$ and $c=1.5s$, we have $2d=1.5s$, so $2(3s)\stackrel{(1)}=1.5s$ since $d=3s$. Now $6s=1.5s$ by $(1)$, which can only happen if $s=0$. Thus $s=d=c=0$.
A: $$d = 3s \tag{1}$$
$$c = \frac{3}{2} s\tag{2}$$
$$c = 2d\tag{3}$$
From $(2)$ and $(3)$, you get
$$2d = \frac{3}{2}s\tag{4}$$
From $(1)$ and $(4)$, you get
$$2\times(3s) = \frac{3}{2} s \tag{5}$$
$(5)$ can be simplified to
$$\left(6-\frac{3}{2}\right)s=0\tag{6}$$
Since $\left(6-\frac{3}{2}\right)\not = 0$, $(6)$ gives you
$$s=0\tag{7}$$
$(1)$ and $(7)$ give you
$$d=0\tag{8}$$
$(3)$ and $(8)$ give you
$$c=0\tag{9}$$
